Binomial Theorem: Class 11 Maths Chapter 7 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

55 min
28 practice questions
17 MCQs

Is chapter mein aap kya seekhenge

Chhote n ke liye (a + b)ⁿ ke expansions mein patterns pehchano aur neighbours ko jodkar Pascal's triangle ki rows banao.
Samjhao ki (a + b)ⁿ ke coefficients combinations ⁿCᵣ kyun hote hain, aur binomial theorem ko poora aur sigma notation mein likho.
Pascal's rule ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ ka use karke binomial theorem ko mathematical induction se prove karo.
(x − y)ⁿ, (1 + x)ⁿ aur (1 − x)ⁿ ko expand karo, aur nikaalo ki binomial coefficients ka sum 2ⁿ hota hai aur unka alternating sum 0 hota hai.
Aise binomials expand karo jinke terms mein reciprocal powers hon, aur trinomials ko binomials ki tarah group karke expand karo.
Theorem ka use karke 10, 100 ya 1000 ke paas wale numbers ki powers nikaalo, badi powers ko numbers se compare karo aur 1 ke paas wale numbers ki powers approximate karo.
(1 + m)ⁿ expand karke divisibility aur remainder wale results prove karo, jaise ki 100, 11ⁿ − 10n − 1 ko divide karta hai.

NCERT Mathematics Class 11, Chapter 7, pp. 126–134 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

7.1 Introduction

Tum saalon se $(a + b)^{2}$ aur $(a - b)^{2}$ ka use kar rahe ho, aksar shortcuts ki tarah: $103^{2} = (100 + 3)^{2} = 10000 + 600 + 9 = 10609$ , bina lambi multiplication ke. Cube identities third powers ke liye yahi kaam karti hain. Lekin higher powers ki kahani alag hai. ₹10,000 ka fixed deposit 3% saalana par, har saal compound hokar, chhe saal baad $10000 \times (1.03)^{6}$ rupaye ka ho jaata hai, aur haath se $1.03$ ko chhe baar khud se multiply karna dheema hai aur ismein galti hona aasaan hai.

Binomial (द्विपद) do terms ka sum hota hai, jaise $a + b$ , $2 x - 5$ ya $1 + 0.03$ . Binomial theorem kisi bhi positive integer $n$ ke liye $(a + b)^{n}$ ko term by term likh deta hai, aur coefficients seedhe padhe ja sakte hain. Yeh chapter sirf positive integer indices ki baat karta hai. Isaac Newton ne 1660s mein is theorem ko negative aur fractional indices tak badhaya; woh version aage ke courses mein aata hai.

Shuru karte hain us identity ki picture se jo tumhe sabse achhi tarah pata hai. Fig. 7.1 side $a + b$ wale square ko aise pieces mein kaatta hai jinke areas jodne par $(a + b)^{2}$ milta hai.

3/3Do pieces barabar hain, isliye $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .

Fig. 7.1

(a + b)^{2}

ka area model: do barabar

a b

pieces se beech ka coefficient 2 milta hai.

Ek bada square jiski upar aur left wali sides dono ek length a aur ek length b mein bati hui hain. Pehle step mein square poora hai aur uska area (a + b) squared likha hai. Doosre step mein dashed cuts ise chaar pieces mein baant dete hain: a squared area ka ek square, ab area ke do rectangles, aur b squared area ka ek chhota square, aur saath mein chaaron areas ka sum likha hai. Teesre step mein dono barabar rectangles outline kiye gaye hain, aur text (a + b) squared = a squared + 2ab + b squared deta hai, coefficients 1, 2, 1 ke saath.

Edge $a + b$ wale cube ko isi tarah kaatne par ek $a^{3}$ block, teen $a^{2} b$ slabs, teen $a b^{2}$ rods aur ek $b^{3}$ cube milta hai, isiliye $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ . Coefficients 1, 2, 1 aur 1, 3, 3, 1 us pattern ke pehle clues hain jo yeh chapter kholta hai.

Quick check

2 mein se 1

a = 100

lekar

(a - b)^{2}

97^{2}

nikaalo.

Answer dekho

Answer 9409

97^{2} = (100 - 3)^{2} = 10000 - 600 + 9 = 9409

. Beech wala term bhoolne par

10000 + 9 = 10009

aata hai, jo sahi nahi ho sakta kyunki

97^{2} < 100^{2}

7.2 Positive Integral Indices ke liye Binomial Theorem

Ek-ek power karke multiply karne se pehle kuch expansions milte hain (zeroth power ke liye $a + b \neq 0$ chahiye). Har line apne upar wali line ko ek baar aur $a + b$ se multiply karke aati hai:

\begin{aligned} (a + b)^{0} & = 1 \\ (a + b)^{1} & = a + b \\ (a + b)^{2} & = a^{2} + 2 a b + b^{2} \\ (a + b)^{3} & = a^{3} + 3 a^{2} b \\ + 3 a b^{2} + b^{3} \\ (a + b)^{4} & = a^{4} + 4 a^{3} b \\ + 6 a^{2} b^{2} \\ + 4 a b^{3} + b^{4} \end{aligned}

(7.1)

Aakhri line lo. $(a + b)^{4} = (a + b)^{3} (a + b)$ , isliye hum $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ ko ek baar $a$ se aur ek baar $b$ se multiply karte hain aur like terms collect karte hain: $a^{4} + (3 + 1) a^{3} b + (3 + 3) a^{2} b^{2} + (1 + 3) a b^{3} + b^{4}$ . Sums $3 + 1$ , $3 + 3$ aur $1 + 3$ par nazar rakho; yeh Pascal's triangle mein wapas aayenge.

(7.1) ki har line mein teen patterns sahi rehte hain:

Terms ki sankhya index se ek zyada hoti hai: $(a + b)^{4}$ mein paanch terms hain.
Ek term se agle term tak, $a$ ki power 1 kam hoti hai aur $b$ ki power 1 badhti hai.
Har term ke andar dono powers ka sum index hota hai; $6 a^{2} b^{2}$ mein, $2 + 2 = 4$ .

Fig. 7.2 $(a + b)^{4}$ ke har term ko uske chaar letters ke stack ki tarah banakar patterns dikhata hai.

4/4Har column 4 cells ooncha hai: powers ka sum index ke barabar hai.

Fig. 7.2

(a + b)^{4}

ka har term chaar letters ka product hai, isliye uski dono powers ka sum hamesha 4 hota hai.NCERT Observations (i)–(iii) · p. 126

Chaar square cells ke paanch columns ek doosre ke bagal mein khade hain, aur unke neeche terms a to the fourth, 4a cubed b, 6a squared b squared, 4a b cubed aur b to the fourth likhe hain. Left se right padhne par columns mein chaar a cells; teen a cells aur ek b cell; dono ke do-do; ek a cell aur teen b cells; aur chaar b cells hain, aur a cells neeche hain. Pehle step mein cells saade hain. Doosre step mein a cells shade hain aur numbers 4, 3, 2, 1, 0 ki ek row a ki power batati hai. Teesre step mein dashed b cells shade hain aur doosri row 0, 1, 2, 3, 4 b ki power batati hai. Chauthe step mein ek double arrow har column ko 4 cells ooncha dikhata hai, is note ke saath ki har term mein powers ka sum 4 hai.

Pascal's triangle

Ab letters hata do aur sirf coefficients rakho, har index ke liye ek row, aur har row upar wali row ke neeche beech mein. Indices 0 se 4 tak ki rows hain 1; 1, 1; 1, 2, 1; 1, 3, 3, 1; aur 1, 4, 6, 4, 1. Hum rows 0 se ginte hain, isliye index $n$ wali row ko row $n$ kehte hain.

Do rules har row ko pichhli row se banate hain. Har row 1 se shuru aur 1 par khatam hoti hai. Row ke andar ka har number uske theek upar wale do numbers ka sum hota hai, jo uske upar-left aur upar-right mein hain; yahi $(a + b)^{4}$ banate waqt wala $3 + 1$ , $3 + 3$ , $1 + 3$ ka hisaab hai. Fig. 7.3 in do rules se triangle ko badhata hai.

4/4Dono edges par sab 1s hain, aur har row beech ke around symmetric hai.

Fig. 7.3Har row 1 se shuru aur 1 par khatam hoti hai, aur andar ka har number upar wale do numbers ka sum hai.NCERT Fig. 7.1, Fig. 7.2 · p. 127

Hexagonal cells ka ek triangle, har index ke liye ek row, upar row 0 se neeche row 6 tak. Pehle step mein sirf rows 0 se 2 dikhti hain: 1; 1, 1; 1, 2, 1. Doosre step mein rows 3 aur 4 aati hain; row 3 ke dono 3s shade hain, unse neeche wale 6 tak arrows jaate hain, aur notes kehte hain 3 + 3 = 6 aur ki andar ka har number upar wale do ka sum hai. Teesre step mein rows 5 aur 6 judti hain: 1, 5, 10, 10, 5, 1 aur 1, 6, 15, 20, 15, 6, 1. Chauthe step mein beech mein ek dashed vertical axis neeche tak jaata hai, dono tirchhe edges ke cells shade hain aur sab mein 1 hai, aur notes batate hain ki har row dono taraf se same padhi jaati hai.

Blaise Pascal (1623–1662), ek French mathematician, ne is array ke baare mein itne facts prove kiye ki ab yeh unke naam se jaana jaata hai: Pascal's triangle. Unse sadiyon pehle, Indian scholars ne ise Meru-prastāra (मेरु प्रस्तार) ke roop mein banaya tha, jab woh Pingala ke bataye poetic metres ke patterns gin rahe the; chapter ke end mein note yeh kahani batata hai.

ExampleNCERT p. 127

Triangle ki ek row se expand karna

Pascal's triangle ka use karke $(2 x + y)^{6}$ expand karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Is method ki ek kamzori hai. $(2 x + y)^{15}$ expand karne ke liye pehle tumhe row 15 tak saari rows chahiye, aur kisi shuru ki row mein ek galti uske neeche ki har row bigaad deti hai. Humein row $n$ ka aisa formula chahiye jo upar ki rows par depend na kare.

Chapter 6 ke combinations yeh formula dete hain. Kisi non-negative integer $n$ aur $0 \leq r \leq n$ ke liye, factorials $n! = 1 \times 2 \times \dots \times n$ aur $0! = 1$ ke saath,

\begin{aligned} ^{n} C_{r} & = \frac{n!}{r! (n - r)!} \\ ^{n} C_{0} & =^{n} C_{n} = 1 \end{aligned}

(7.2)

jahan

$^{n} C_{r}$: $n$ objects mein se $r$ objects chunne ke tareekon ki sankhya

Ise row 4 par test karo: $^{4} C_{0},^{4} C_{1},^{4} C_{2},^{4} C_{3},^{4} C_{4}$ hain $1, 4, 6, 4, 1$ , bilkul wahi row. Row 8, rows 5 se 7 ke bina seedhe formula se nikaali gayi, hai 1, 8, 28, 56, 70, 56, 28, 8, 1. General mein, Pascal's triangle ki row $n$ hai $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ .

Ek counting formula yeh coefficients kyun dega? Fig. 7.4 $(a + b)^{3}$ ke liye is sawaal ka jawab deta hai.

3/3Kaun se $r$ brackets $b$ denge, yeh chunne ke $^{3} C_{r}$ tareeke hain: 1, 3, 3, 1.

Fig. 7.43 brackets mein se

r

brackets se

b

chunna

^{3} C_{r}

tareekon se ho sakta hai, isliye

a^{3 - r} b^{r}

ka coefficient

^{3} C_{r}

hai.NCERT Fig. 7.3 · p. 128

Upar product (a + b)(a + b)(a + b) hai. Pehle step mein har bracket se ek letter chunne ke aath tareeke ek row mein likhe hain: aaa, aab, aba, baa, abb, bab, bba aur bbb, note 2 × 2 × 2 = 8 choices ke saath; har b bold mein chhapa hai. Doosre step mein words chaar columns mein khisak jaate hain jinke headings a cubed, a squared b, a b squared aur b cubed hain, is hisaab se ki unmein kitne b letters hain, aur neeche 1 way, 3 ways, 3 ways aur 1 way likha hai. Teesre step mein counts par combinations 3C0, 3C1, 3C2 aur 3C3 ke labels hain.

Yahi reasoning kisi bhi index ke liye kaam karti hai. $(a + b)^{n}$ ko multiply karne ka matlab hai $n$ brackets mein se har ek se ek letter har possible tareeke se chunna aur products ko jodna. Koi product $a^{n - r} b^{r}$ ke barabar tabhi hota hai jab $b$ theek $r$ brackets se aaya ho, aur woh $r$ brackets $^{n} C_{r}$ tareekon se chune ja sakte hain. Isliye $a^{n - r} b^{r}$ $^{n} C_{r}$ baar aata hai, aur uska coefficient $^{n} C_{r}$ hai. Jaise row 6, combinations ke roop mein:

\begin{aligned} (a + b)^{6} \\ =^{6} C_{0} a^{6} +^{6} C_{1} a^{5} b \\ +^{6} C_{2} a^{4} b^{2} +^{6} C_{3} a^{3} b^{3} \\ +^{6} C_{4} a^{2} b^{4} +^{6} C_{5} a b^{5} \\ +^{6} C_{6} b^{6} \end{aligned}

(7.3)

7.2.1 Kisi bhi positive integer $n$ ke liye binomial theorem

Chunne wala argument har index ke liye kaam karta hai, aur isse general statement milta hai:

Binomial theorem

\begin{aligned} (a + b)^{n} \\ =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b \\ +^{n} C_{2} a^{n - 2} b^{2} + \dots \\ +^{n} C_{n - 1} a b^{n - 1} \\ +^{n} C_{n} b^{n} \end{aligned}

(7.4)

Counting argument pehle se hi ek convincing reason hai, lekin standard proof, jo examiners expect karte hain, mathematical induction use karta hai. Is book mein ab induction ka apna alag chapter nahi hai, isliye yahan uska idea chhote mein diya hai.

3/3Dono steps milkar har positive integer $n$ ke liye $P (n)$ dete hain.

Fig. 7.5Pehla domino girta hai, aur har girta domino agle ko gira deta hai, isliye sab gir jaate hain.

Chhe dominoes, 1 se 6 tak numbered, ek line par row mein khade hain, aur unke baad dots hain jo dikhate hain ki row aage chalti hai. Pehle step mein ek arrow pehle domino ko dhakka deta hai, jo shaded hai, is note ke saath: pehle, n = 1 sahi hai. Doosre step mein dominoes 1 aur 2 gir chuke hain aur agla shaded hai, is note ke saath ki case k, case k + 1 ko gira deta hai. Teesre step mein har domino gir chuka hai, is note ke saath: isliye har case girta hai.

Proof sigma notation mein sabse chhota hai. Symbol $\sum_{r = 0}^{n}$ ka matlab hai: expression ko baari-baari $r = 0, 1, 2, \dots, n$ ke liye likho aur results jodo. $a^{0}$ aur $b^{0}$ ko 1 padhte hue, theorem ban jaata hai

\begin{aligned} (a + b)^{n} \\ = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r} \end{aligned}

(7.5)

Proof ko Chapter 6 ka ek fact bhi chahiye, Pascal's rule, jo bilkul triangle ka upar-ke-do-jodo wala rule hai:

Pascal's rule

\begin{aligned} ^{n} C_{r} +^{n} C_{r - 1} \\ =^{n + 1} C_{r} \end{aligned}

(7.6)

jahan

$r$: koi bhi whole number jahan $1 \leq r \leq n$

TheoremBinomial theorem

Har positive integer $n$ aur kinhi bhi numbers $a$ aur $b$ ke liye, expansion (7.4) sahi hai; sigma form mein, $(a + b)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}$ .

Proof. Maan lo $P (n)$ statement (7.5) hai. Hum pehla case check karte hain aur phir dikhate hain ki har case agle case ko le aata hai.

Pehla case. $n = 1$ ke liye right side $^{1} C_{0} a +^{1} C_{1} b = a + b$ hai, jo $(a + b)^{1}$ hai. Isliye $P (1)$ sahi hai.

$k$ se $k + 1$ tak. Ek positive integer $k$ fix karo aur maan lo ki $P (k)$ sahi hai:

\begin{aligned} (a + b)^{k} \\ = \sum_{r = 0}^{k}^{k} C_{r} a^{k - r} b^{r} \end{aligned}

Dono sides ko $a + b$ se multiply karo. $a$ se multiply karne par $a$ ki har power ek badh jaati hai; $b$ se multiply karne par $b$ ki har power ek badh jaati hai:

\begin{aligned} (a + b)^{k + 1} \\ = \sum_{r = 0}^{k}^{k} C_{r} a^{k + 1 - r} b^{r} \\ + \sum_{r = 0}^{k}^{k} C_{r} a^{k - r} b^{r + 1} \end{aligned}

Doosre sum mein $s = r + 1$ rakho. Tab $s$ 1 se $k + 1$ tak chalta hai aur general term $^{k} C_{s - 1} a^{k + 1 - s} b^{s}$ hai. $s$ ko phir se $r$ naam dene par, ab dono sums mein same powers $a^{k + 1 - r} b^{r}$ hain.

Pehle sum se $r = 0$ wala term aur doosre sum se $r = k + 1$ wala term bahar nikaalo, phir jo bacha hai unke pairs banao:

\begin{aligned} (a + b)^{k + 1} \\ = a^{k + 1} + b^{k + 1} \\ + \sum_{r = 1}^{k} c_{r} a^{k + 1 - r} b^{r} \end{aligned}

jahan $c_{r} =^{k} C_{r} +^{k} C_{r - 1}$ . Pascal's rule (7.6) se, $c_{r} =^{k + 1} C_{r}$ . Saath hi $a^{k + 1} =^{k + 1} C_{0} a^{k + 1}$ aur $b^{k + 1} =^{k + 1} C_{k + 1} b^{k + 1}$ , kyunki dono coefficients 1 hain. Isliye

\begin{aligned} (a + b)^{k + 1} \\ = \sum_{r = 0}^{k + 1}^{k + 1} C_{r} a^{k + 1 - r} b^{r} \end{aligned}

jo bilkul $P (k + 1)$ hai. Pehla case sahi hai aur har case agle ko zaroori bana deta hai, isliye principle of mathematical induction se $P (n)$ bina kisi exception ke $n = 1, 2, 3, \dots$ ke liye sahi hai.

Fig. 7.6 $k = 4$ ke liye proof ka core saade numbers se dikhata hai. $a (a + b)^{4}$ aur $b (a + b)^{4}$ ke coefficients same row hain, bas ek jagah khiske hue, aur columns jodne se row 5 banti hai.

3/3Har column ka sum $^{4} C_{r} +^{4} C_{r - 1} =^{5} C_{r}$ hai, jisse row 5 milti hai.

Fig. 7.6

a + b

se multiply karna row 4 ko uski ek khiski hui copy se jodta hai, aur row 5 milti hai: Pascal's rule kaam karte hue.NCERT p. 129

Chhe columns ke upar powers a to the fifth, a to the fourth b, a cubed b squared, a squared b cubed, a b to the fourth aur b to the fifth likhi hain. Pehle step mein row 1, 4, 6, 4, 1 pehle paanch columns bharti hai, label times a ke saath. Doosre step mein row ki doosri copy, label times b ke saath, ek column aur right khisak kar columns do se chhe bharti hai. Teesre step mein dono rows ke neeche ek line kheenchi jaati hai aur column sums 1, 5, 10, 10, 5, 1 dikhte hain, 6 aur 4 wala column highlight hai aur note 6 + 4 = 10 hai.

ExampleNCERT p. 129

Theorem se pehla expansion

Binomial theorem ka use karke $(y + 3)^{5}$ expand karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Definition

Binomial coefficients

Numbers $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ jo $(a + b)^{n}$ ke terms ko multiply karte hain. Pascal's triangle ki row $n$ inhe order mein likhti hai.

Example $n = 4$ ke liye yeh 1, 4, 6, 4, 1 hain.

Theorem ko dhyaan se padhne par kisi bhi expansion ke liye ek checklist milti hai:

Terms ki sankhya. $(a + b)^{n}$ mein $n + 1$ terms hote hain, har $r = 0, 1, \dots, n$ ke liye ek.
Powers. Left se right, $a$ ki power $n$ se 0 tak utarti hai jabki $b$ ki power 0 se $n$ tak chadhti hai.
Degree. Har term mein dono powers ka sum $n$ hota hai, kyunki $(n - r) + r = n$ .
Symmetry. Kyunki $^{n} C_{r} =^{n} C_{n - r}$ , coefficients dono taraf se same padhe jaate hain.

InteractiveBinomial lab

Do tabs. Triangle tab mein ek slider Pascal's triangle ki rows 0 se 10 dikhata hai; kisi entry par tap karne se woh aur uske upar wali do entries highlight hoti hain, aur ek panel uski row aur position, factorial formula se uski value, upar wale do numbers ka sum, uski mirror-image entry aur uski row ka sum 2 ki power ki tarah deta hai. Expand tab mein ek slider index n ko 1 se 10 tak set karta hai. Letters mode mein buttons whole numbers p aur q ko −5 se 5 tak set karte hain aur ek table har r ke liye binomial coefficient, powers ka product aur (px + qy) to the n ka simplified term deti hai, phir poora expansion aur ek check ki coefficients ka sum (p + q) to the n hai. Numbers mode mein A 10, 100 ya 1000 hai aur B −9 se 9 tak ka whole number hai; table (A + B) to the n ka har term running total ke saath deti hai jo exact value par khatam hota hai.

Fig. 7.7Pascal's triangle ki kisi bhi entry par tap karo aur use

^{n} C_{r}

ki tarah aur upar wale do ke sum ki tarah dekho, ya

(p x + q y)^{n}

aur

(100 - 3)^{n}

jaisi powers ko term by term expand karo.

Quick check

4 mein se 1

(3 p - 2 q)^{11}

ke expansion mein kitne terms hain?

Answer dekho

Answer 12

Index se ek zyada:

11 + 1 = 12

. 11 answer dene mein

r = 0

wala term chhoot jaata hai.

7.2.2 Kuch special cases

Theorem kinhi bhi numbers $a$ aur $b$ ke liye sahi hai, isliye hum ismein special values daal sakte hain. Teen choices baar-baar use hoti hain.

(i) Ek difference. $a = x$ aur $b = - y$ rakho. Kyunki even $r$ ke liye $(- y)^{r}$ $y^{r}$ hai aur odd $r$ ke liye $- y^{r}$ , har term ka size wahi rehta hai lekin signs alternate hote hain:

\begin{aligned} (x - y)^{n} \\ =^{n} C_{0} x^{n} -^{n} C_{1} x^{n - 1} y \\ +^{n} C_{2} x^{n - 2} y^{2} - \dots \\ + (- 1)^{n}^{n} C_{n} y^{n} \end{aligned}

(7.7)

ExampleNCERT p. 130

Alternating signs

$(x - 3 y)^{4}$ expand karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

(ii) Ek plus kuch. $a = 1$ aur $b = x$ rakho. 1 ki har power 1 hai, isliye sirf $x$ ki powers bachti hain:

\begin{aligned} (1 + x)^{n} \\ =^{n} C_{0} +^{n} C_{1} x \\ +^{n} C_{2} x^{2} + \dots \\ +^{n} C_{n} x^{n} \end{aligned}

(7.8)

Yeh form approximations aur divisibility proofs mein zyadatar kaam karta hai, jaisa neeche ke examples dikhate hain. Ismein $x = 1$ rakhne se triangle ki poori row jud jaati hai:

\begin{aligned} ^{n} C_{0} +^{n} C_{1} +^{n} C_{2} \\ + \dots +^{n} C_{n} = 2^{n} \end{aligned}

(7.9)

Iska ek counting reason bhi hai. $^{n} C_{r}$ , $n$ elements wale set ke $r$ elements wale subsets ginta hai, isliye left side uske saare subsets ginti hai. Subset banane ka matlab hai har element ke liye in ya out decide karna, jisse $2 \times 2 \times \dots \times 2 = 2^{n}$ choices milti hain.

(iii) Ek minus kuch. $a = 1$ aur $b = - x$ rakho:

\begin{aligned} (1 - x)^{n} \\ =^{n} C_{0} -^{n} C_{1} x \\ +^{n} C_{2} x^{2} - \dots \\ + (- 1)^{n}^{n} C_{n} x^{n} \end{aligned}

(7.10)

$x = 1$ ke saath left side har positive integer $n$ ke liye $0^{n} = 0$ hai, isliye alternating signs wale coefficients poori tarah cancel ho jaate hain:

\begin{aligned} ^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - \dots \\ + (- 1)^{n}^{n} C_{n} = 0 \end{aligned}

(7.11)

Fig. 7.8 dono facts ko pehli rows par check karta hai. Sirf row 0 doosre fact ko todti hai, kyunki $(1 - 1)^{0} = 1$ ; woh row bahar hai, kyunki $n$ positive integer hai.

3/3Signs $+, -, +, -, \dots$ ke saath har total 0 hai, row 0 ko chhodkar.

Fig. 7.8Row

n

ka sum

2^{n}

hai; alternating signs ke saath, row 0 ke baad har row ka sum 0 hai.NCERT p. 131

Pascal's triangle ki rows 0 se 5 left-aligned staircase ki tarah likhi hain, har line mein ek row, upar 1 se neeche 1, 5, 10, 10, 5, 1 tak. Pehle step mein sirf rows dikhti hain. Doosre step mein sum heading wala ek shaded column har row ka total deta hai: 1, 2, 4, 8, 16 aur 32, har baar double. Teesre step mein har row ke har doosre number ke aage minus signs aate hain, aur plus minus plus minus heading wala doosra column in alternating signs ke saath totals deta hai: sabse upar wali row ke liye 1 aur baaki har row ke liye 0.

(7.9) aur (7.11) ko jodne par odd positions wale coefficients cancel ho jaate hain, aur subtract karne par even positions wale. Isliye row ke har aadhe hisse mein total ka theek aadha hota hai:

\begin{aligned} ^{n} C_{0} +^{n} C_{2} + \dots \\ =^{n} C_{1} +^{n} C_{3} + \dots \\ = 2^{n - 1} \end{aligned}

(7.12)

Quick check

3 mein se 1

(2 - x)^{7}

mein

x^{3}

ka coefficient hai

560

- 560

- 280

- 35

Answer dekho

Answer (B) $- 560$

Term hai

^{7} C_{3} 2^{4} (- x)^{3} = 35 \times 16 \times (- x^{3}) = - 560 x^{3}

. Sign

- x

ki odd power se aata hai;

- 280

mein

2^{7 - 3} = 2^{4}

ki jagah

2^{3}

use hua hai.

Theorem ka use

Example 7.1NCERT Example 1 · p. 131

Reciprocal power wala binomial

${(3 x - \frac{1}{x^{2}})}^{4}$ expand karo, jahan $x \neq 0$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 7.2NCERT Example 2 · p. 131

Lambi multiplication ke bina fourth power

$97^{4}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 7.3NCERT Example 3 · p. 131

Ek bahut badi power ko ek number se compare karna

Kaun bada hai, $(1.002)^{5000}$ ya 10?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Do-term wala bound safe kyun hai, aur yeh kitni jaldi accuracy khota hai? Fig. 7.9 $(1.1)^{n}$ ko uske pehle do aur pehle teen terms se compare karta hai.

3/3 $^{n} C_{2} (0.1)^{2}$ jodne se value paas aati hai, lekin phir bhi neeche rehti hai.

Fig. 7.9Hataye gaye saare terms positive hain, isliye har truncation

(1.1)^{n}

se neeche rehta hai, aur gap

n

ke saath badhta hai.NCERT p. 132

Ek graph jismein across n 0 se 20 tak aur upar values 0 se 7 tak hain. Pehle step mein dots har whole number n par 1.1 to the power n dikhate hain, jo 1 se lagbhag 6.7 tak badhte hain aur upar ki taraf mudte hain. Doosre step mein pehle do terms, 1 + 0.1n, ki ek dashed seedhi line judti hai; yeh usi point se shuru hoti hai lekin dots ke neeche rehti hai aur n = 20 par sirf 3 tak pahunchti hai, aur n = 16 par ek double arrow gap dikhata hai, label gap greater than 0 ke saath. Teesre step mein pehle teen terms ka ek dotted curve judta hai; yeh line aur dots ke beech rehta hai, chhote n ke liye dots ke paas, aur n = 20 par 4.9 tak pahunchta hai.

Example 7.4NCERT Example 4 · p. 132

11 ki power ke aakhri do digits

Dikhao ki har positive integer $n$ ke liye $11^{n} - 10 n - 1$ , 100 se divisible hai. Isse $11^{7}$ ke aakhri do digits nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Wapas fixed deposit par

Expansion ke pehle teen terms use karke $10000 \times (1.03)^{6}$ estimate karo, jo Section 7.1 wala deposit hai. Estimate kitna achha hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Teen terms ko do ki tarah lena

$(1 + x + x^{2})^{3}$ expand karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Pehle teen terms use karke,

(1.01)^{8}

lagbhag hai

1.0800

1.0882

1.0280

1.0828

Answer dekho

Answer (D) $1.0828$

1 + 8 (0.01) + 28 (0.01)^{2} = 1 + 0.08 + 0.0028 = 1.0828

; paanch decimal places tak exact value 1.08286 hai. 1.0800 do terms ke baad ruk jaata hai.

Practice · 14 questions

Practice 7.1: answers ke saath

Short answerEasyNCERT Exercise 7.1 Q1
$(1 - 3 x)^{5}$ expand karo.
Solution dikhao
Answer $1 - 15 x + 90 x^{2} - 270 x^{3} + 405 x^{4} - 243 x^{5}$
(7.10) ko $x$ ki jagah $3 x$ lekar use karo. $- 3 x$ ke $r$ factors wala term $^{5} C_{r} (- 3 x)^{r}$ hai:
$\begin{aligned} (1 - 3 x)^{5} \\ = 1 - 5 (3 x) + 10 (9 x^{2}) \\ - 10 (27 x^{3}) + 5 (81 x^{4}) \\ - 243 x^{5} \end{aligned}$
Isliye $(1 - 3 x)^{5} = 1 - 15 x + 90 x^{2} - 270 x^{3} + 405 x^{4} - 243 x^{5}$ . $x = 1$ se check karo: sum $- 32 = (- 2)^{5}$ hai.
Short answerMediumNCERT Exercise 7.1 Q2
${(\frac{x}{3} + \frac{3}{x})}^{4}$ expand karo, $x \neq 0$ .
Solution dikhao
Answer $\frac{x^{4}}{81} + \frac{4 x^{2}}{9} + 6 + \frac{36}{x^{2}} + \frac{81}{x^{4}}$
$a = \frac{x}{3}$ , $b = \frac{3}{x}$ aur row 4 = 1, 4, 6, 4, 1 ke saath:
$\begin{aligned} r = 0 & : \frac{x^{4}}{81} \\ r = 1 & : 4 \cdot \frac{x^{3}}{27} \cdot \frac{3}{x} = \frac{4 x^{2}}{9} \\ r = 2 & : 6 \cdot \frac{x^{2}}{9} \cdot \frac{9}{x^{2}} = 6 \\ r = 3 & : 4 \cdot \frac{x}{3} \cdot \frac{27}{x^{3}} = \frac{36}{x^{2}} \\ r = 4 & : \frac{81}{x^{4}} \end{aligned}$
Paanchon terms jodo. Beech wala term $x$ se free hai kyunki wahan $x$ aur $\frac{1}{x}$ barabar baar aate hain.
Short answerEasyNCERT Exercise 7.1 Q3
$(2 x - 5)^{4}$ expand karo.
Solution dikhao
Answer $16 x^{4} - 160 x^{3} + 600 x^{2} - 1000 x + 625$
$\begin{aligned} (2 x - 5)^{4} \\ = (2 x)^{4} - 4 (2 x)^{3} (5) \\ + 6 (2 x)^{2} (5^{2}) \\ - 4 (2 x) (5^{3}) + 5^{4} \end{aligned}$
$= 16 x^{4} - 160 x^{3} + 600 x^{2} - 1000 x + 625$ . $x = 1$ se check karo: $16 - 160 + 600 - 1000 + 625 = 81 = (- 3)^{4}$ .
Short answerMediumNCERT Exercise 7.1 Q4
${(\frac{x}{2} - \frac{1}{x})}^{6}$ expand karo, $x \neq 0$ .
Solution dikhao
Answer $\frac{x^{6}}{64} - \frac{3 x^{4}}{16} + \frac{15 x^{2}}{16} - \frac{5}{2} + \frac{15}{4 x^{2}} - \frac{3}{x^{4}} + \frac{1}{x^{6}}$
Row 6 hai 1, 6, 15, 20, 15, 6, 1 aur ${(\frac{1}{2})}^{6}$ se ${(\frac{1}{2})}^{0}$ tak $\frac{1}{2}$ ki powers hain $\frac{1}{64}, \frac{1}{32}, \frac{1}{16}, \frac{1}{8}, \frac{1}{4}, \frac{1}{2}, 1$ . Multiply karke aur alternating signs lagakar:
$\begin{aligned} \frac{1}{64}, - \frac{6}{32}, \frac{15}{16}, - \frac{20}{8}, \\ \frac{15}{4}, - \frac{6}{2}, 1 \end{aligned}$
Yeh $x^{6}, x^{4}, x^{2}, x^{0}, x^{- 2}, x^{- 4}, x^{- 6}$ ko multiply karte hain. Simplify karne par $\frac{6}{32} = \frac{3}{16}$ , $\frac{20}{8} = \frac{5}{2}$ aur $\frac{6}{2} = 3$ , jisse answer milta hai.
Short answerEasyNCERT Exercise 7.1 Q5
${(x - \frac{1}{x})}^{5}$ expand karo, $x \neq 0$ .
Solution dikhao
Answer $x^{5} - 5 x^{3} + 10 x - \frac{10}{x} + \frac{5}{x^{3}} - \frac{1}{x^{5}}$
$- \frac{1}{x}$ ke $r$ factors wala term $^{5} C_{r} (- 1)^{r} x^{5 - 2 r}$ hai. Row 5 = 1, 5, 10, 10, 5, 1 aur alternating signs ke saath, expansion $x^{5} - 5 x^{3} + 10 x - 10 x^{- 1} + 5 x^{- 3} - x^{- 5}$ hai.
NumericalEasyNCERT Exercise 7.1 Q6
Binomial theorem ka use karke $95^{3}$ nikaalo.
Solution dikhao
$\begin{aligned} (100 - 5)^{3} \\ = 1 000 000 - 150 000 \\ + 7 500 - 125 \end{aligned}$
$= 857 375$ .
NumericalEasyNCERT Exercise 7.1 Q7
Binomial theorem ka use karke $103^{4}$ nikaalo.
Solution dikhao
$\begin{aligned} (100 + 3)^{4} \\ = 100 000 000 + 12 000 000 \\ + 540 000 + 10 800 + 81 \end{aligned}$
$= 112 550 881$ .
NumericalEasyNCERT Exercise 7.1 Q8
Binomial theorem ka use karke $1001^{3}$ nikaalo.
Solution dikhao
$\begin{aligned} (1000 + 1)^{3} & = 10^{9} + 3 \times 10^{6} \\ + 3 \times 10^{3} + 1 \end{aligned}$
$= 1 003 003 001$ .
NumericalEasyNCERT Exercise 7.1 Q9
Binomial theorem ka use karke $999^{4}$ nikaalo.
Solution dikhao
$\begin{aligned} (1000 - 1)^{4} \\ = 10^{12} - 4 \times 10^{9} \\ + 6 \times 10^{6} - 4 \times 10^{3} + 1 \end{aligned}$
$= 996 005 996 001$ .
Quick answerMediumNCERT Exercise 7.1 Q10
Bina calculator ke decide karo ki kaun bada hai: $(1.05)^{400}$ ya 20.
20 $(1.05)^{400}$
Solution dikhao
$(1.05)^{400} = 1 + 400 (0.05) + (positive terms) > 1 + 20 = 21$ , isliye $(1.05)^{400} > 20$ . (Iska asli size lagbhag $3 \times 10^{8}$ hai.)
11
Short answerMediumNCERT Exercise 7.1 Q11
Fifth powers ke differences.
1. $(x + y)^{5} - (x - y)^{5}$ ko expand aur simplify karo.
  Solution dikhao
  Answer $2 (5 x^{4} y + 10 x^{2} y^{3} + y^{5})$
  $(x - y)^{5}$ mein $y$ ki odd powers wale terms ka sign badalta hai, isliye difference mein even-power terms cancel ho jaate hain aur odd-power terms double ho jaate hain: $2 (^{5} C_{1} x^{4} y +^{5} C_{3} x^{2} y^{3} +^{5} C_{5} y^{5}) = 10 x^{4} y + 20 x^{2} y^{3} + 2 y^{5}$ .
2. Isse $(2 + \sqrt{3})^{5} - (2 - \sqrt{3})^{5}$ nikaalo.
  Solution dikhao
  Answer $418 \sqrt{3}$
  $x = 2$ aur $y = \sqrt{3}$ rakho, to $y^{3} = 3 \sqrt{3}$ aur $y^{5} = 9 \sqrt{3}$ :
  $\begin{aligned} 2 (80 \sqrt{3} + 120 \sqrt{3} + 9 \sqrt{3}) \\ = 418 \sqrt{3} \end{aligned}$
12
Short answerMediumNCERT Exercise 7.1 Q12
Fourth powers ke sums.
1. $(x + 1)^{4} + (x - 1)^{4}$ ko expand aur simplify karo.
  Solution dikhao
  Answer $2 x^{4} + 12 x^{2} + 2$
  Odd-power terms $4 x^{3}$ aur $4 x$ cancel ho jaate hain, aur baaki double ho jaate hain: $2 (x^{4} + 6 x^{2} + 1) = 2 x^{4} + 12 x^{2} + 2$ .
2. Isse $(\sqrt{5} + 1)^{4} + (\sqrt{5} - 1)^{4}$ nikaalo.
  Solution dikhao
  $x = \sqrt{5}$ ke saath: $x^{2} = 5$ aur $x^{4} = 25$ , isliye value $2 (25 + 30 + 1) = 112$ hai.
ProofHardNCERT Exercise 7.1 Q13
Dikhao ki har positive integer $n$ ke liye $5^{n + 1} - 4 n - 5$ , 16 se divisible hai.
Solution dikhao
(7.8) se, $x = 4$ ke saath:
$\begin{aligned} 5^{n + 1} & = 5 (1 + 4)^{n} \\ = 5 (1 + 4 n +^{n} C_{2} 4^{2} \\ + \dots + 4^{n}) \\ = 5 + 20 n + 80 k \end{aligned}$
jahan $k =^{n} C_{2} +^{n} C_{3} 4 + \dots + 4^{n - 2}$ ek integer hai ( $n = 1$ ke liye $k = 0$ ). Tab $5^{n + 1} - 4 n - 5 = 16 n + 80 k = 16 (n + 5 k)$ , jo 16 ka multiple hai. Check: $n = 2$ se $125 - 8 - 5 = 112 = 16 \times 7$ milta hai.
ProofMediumNCERT Exercise 7.1 Q14
Prove karo ki $\sum_{r = 0}^{n} 2^{r}^{n} C_{r} = 3^{n}$ .
Solution dikhao
$(1 + x)^{n} = \sum_{r = 0}^{n}^{n} C_{r} x^{r}$ mein $x = 2$ rakho: left side $3^{n}$ hai aur right side $\sum 2^{r}^{n} C_{r}$ hai. $n = 3$ ke liye: $1 + 6 + 12 + 8 = 27 = 3^{3}$ .

Practice · 6 questions

Chapter 7 par miscellaneous exercises: answers ke saath

1
ProofHardNCERT Miscellaneous Exercise Q1
Ek divisibility pattern.
1. Dikhao ki har positive integer $n$ ke liye $17^{n} - 9^{n}$ , 8 se divisible hai.
  Solution dikhao
  $(9 + 8)^{n} = 9^{n} +^{n} C_{1} 9^{n - 1} \cdot 8 +^{n} C_{2} 9^{n - 2} \cdot 8^{2} + \dots + 8^{n}$ . $9^{n}$ ke baad ke har term mein factor 8 hai, isliye $17^{n} - 9^{n} = 8 (^{n} C_{1} 9^{n - 1} +^{n} C_{2} 9^{n - 2} \cdot 8 + \dots + 8^{n - 1})$ , jo 8 ka integer multiple hai.
2. Samjhao ki yahi idea kaise dikhata hai ki kinhi bhi do alag integers $a$ aur $b$ ke liye $a - b$ , $a^{n} - b^{n}$ ko divide karta hai.
  Solution dikhao
  $a = b + (a - b)$ likho aur $(b + (a - b))^{n} = b^{n} +^{n} C_{1} b^{n - 1} (a - b) + \dots + (a - b)^{n}$ expand karo. $b^{n}$ ke baad ke har term mein $a - b$ hai, isliye $a^{n} - b^{n} = (a - b) \times (an integer)$ .
NumericalMediumNCERT Miscellaneous Exercise Q2
$(\sqrt{3} + 1)^{6} - (\sqrt{3} - 1)^{6}$ ko $k \sqrt{3}$ ke form mein likho. $k$ enter karo.
Solution dikhao
$(x + y)^{6} - (x - y)^{6} = 2 (6 x^{5} y + 20 x^{3} y^{3} + 6 x y^{5})$ . $x = \sqrt{3}$ , $y = 1$ ke saath: $x^{5} = 9 \sqrt{3}$ aur $x^{3} = 3 \sqrt{3}$ , isliye value $2 (54 \sqrt{3} + 60 \sqrt{3} + 6 \sqrt{3}) = 240 \sqrt{3}$ hai. Isliye $k = 240$ .
Short answerHardNCERT Miscellaneous Exercise Q3
${(x + \sqrt{x^{2} - 4})}^{4} + {(x - \sqrt{x^{2} - 4})}^{4}$ ko simplify karo, jahan $x^{2} \geq 4$ .
Solution dikhao
Answer $16 x^{4} - 64 x^{2} + 32$
$\begin{aligned} (x + s)^{4} + (x - s)^{4} \\ = 2 (x^{4} + 6 x^{2} s^{2} + s^{4}) \end{aligned}$
$s^{2} = x^{2} - 4$ aur $s^{4} = x^{4} - 8 x^{2} + 16$ ke saath:
$\begin{aligned} 2 (x^{4} + 6 x^{4} - 24 x^{2} \\ + x^{4} - 8 x^{2} + 16) \\ = 16 x^{4} - 64 x^{2} + 32 \end{aligned}$
$x = 3$ se check karo: $s = \sqrt{5}$ aur $(3 + \sqrt{5})^{4} + (3 - \sqrt{5})^{4} = 752 = 16 (81) - 64 (9) + 32$ .
NumericalMediumNCERT Miscellaneous Exercise Q4
Expansion ke pehle teen terms use karke $(0.98)^{6}$ ko teen decimal places tak approximate karo.
Solution dikhao
$(1 - 0.02)^{6} \approx 1 - 6 (0.02) + 15 (0.02)^{2} = 1 - 0.12 + 0.006 = 0.886$ . Agla term, $- 20 (0.02)^{3} = - 0.00016$ , teesre decimal ko nahi badalta; exact value $0.88584 \dots$ hai
Short answerHardNCERT Miscellaneous Exercise Q5
${(1 + x - \frac{1}{x})}^{3}$ expand karo, $x \neq 0$ .
Solution dikhao
Answer $x^{3} + 3 x^{2} - 5 + \frac{3}{x^{2}} - \frac{1}{x^{3}}$
$(1 + u)^{3} = 1 + 3 u + 3 u^{2} + u^{3}$ , jahan
$\begin{aligned} u^{2} & = x^{2} - 2 + \frac{1}{x^{2}} \\ u^{3} & = x^{3} - 3 x + \frac{3}{x} - \frac{1}{x^{3}} \end{aligned}$
$1 + (3 x - \frac{3}{x}) + (3 x^{2} - 6 + \frac{3}{x^{2}}) + u^{3}$ jodne par $3 x$ aur $\frac{3}{x}$ wale terms cancel ho jaate hain, aur bachta hai $x^{3} + 3 x^{2} - 5 + \frac{3}{x^{2}} - \frac{1}{x^{3}}$ . $x = 1$ se check karo: $1 + 3 - 5 + 3 - 1 = 1 = 1^{3}$ .
Short answerHardNCERT Miscellaneous Exercise Q6
$(x^{2} - 2 x + 2)^{3}$ ko binomial ki tarah likhkar expand karo.
Solution dikhao
Answer $x^{6} - 6 x^{5} + 18 x^{4} - 32 x^{3} + 36 x^{2} - 24 x + 8$
$(v + 2)^{3} = v^{3} + 6 v^{2} + 12 v + 8$ , jahan $v^{2} = x^{4} - 4 x^{3} + 4 x^{2}$ aur $v^{3} = x^{6} - 6 x^{5} + 12 x^{4} - 8 x^{3}$ :
$\begin{aligned} x^{6} - 6 x^{5} + 12 x^{4} - 8 x^{3} \\ + 6 x^{4} - 24 x^{3} + 24 x^{2} \\ + 12 x^{2} - 24 x + 8 \end{aligned}$
Collect karne par: $x^{6} - 6 x^{5} + 18 x^{4} - 32 x^{3} + 36 x^{2} - 24 x + 8$ . $x = 1$ se check karo: sum $1 = 1^{3}$ hai.

Recap

Summary aur key points

Binomial ki har power pichhli power se aati hai: $(a + b)^{n + 1} = (a + b)^{n} (a + b)$ (Section 7.2).
$(a + b)^{n}$ mein $n + 1$ terms hote hain; $a$ ki power $n$ se 0 tak girti hai jab $b$ ki power 0 se $n$ tak badhti hai, aur har term mein dono powers ka sum $n$ hota hai (teen patterns).
Pascal's triangle, yaani Meru-prastāra, coefficients ki list deta hai: har row ke dono ends par 1, aur andar ki har entry upar wali do entries ka sum (Pascal's triangle).
Row $n$ hai $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ ((7.2)): $a^{n - r} b^{r}$ ka coefficient un tareekon ko ginta hai jinse $b$ dene wale $r$ brackets chune ja sakte hain.
Binomial theorem: har positive integer $n$ ke liye $(a + b)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}$ (Theorem), jo Pascal's rule ((7.6)) ke saath induction se prove hota hai.
$(x - y)^{n}$ ke coefficients same hote hain, bas signs alternate hote hain ((7.7)).
$(1 + x)^{n} = \sum^{n} C_{r} x^{r}$ ; $x = 1$ par binomial coefficients ka sum $2^{n}$ hota hai, aur alternating signs ke saath 0 ((7.9), (7.11)).
10, 100 ya 1000 ke paas wale numbers ki powers jaldi nikalti hain, jab number ko ek round number plus ya minus ek chhote number ki tarah likha jaaye (Example 7.2).
$x > 0$ ke liye $(1 + x)^{n}$ ka har term positive hai, isliye pehle kuch terms ek lower bound ya close estimate dete hain (Example 7.3, deposit wala estimate).
Base ko $1 + m$ likhkar expand karne se $100 ∣ 11^{n} - 10 n - 1$ jaise divisibility facts prove hote hain (Example 7.4).

Flash cards · 6

Formulas ek nazar mein

1 / 6

History seNCERT Historical Note · p. 134

Meru-prastāra se Pascal tak

Indian prosody in numbers ke sabse pehle gharon mein se ek tha. Pingala ka Chhandaḥśāstra, poetic metre par do hazaar saal se bhi pehle likha gaya ek granth, poochhta hai ki ek line mein long aur short syllables ke kitne patterns ho sakte hain, aur unmein se kitne mein diye gaye number ke short syllables hote hain. Answers binomial coefficients hain. Lagbhag 10th century mein commentator Halāyudha ne Pingala ke rules ko ek triangular figure se samjhaya, Meru-prastāra ('Mount Meru ki seedhi'), jo paas wale numbers jodkar banta hai.

Yahi triangle lagbhag 1000 CE mein Baghdad mein al-Karajī ke kaam mein dikhta hai, aur China mein Yang Hui ki 1261 ki book mein aur Zhu Shijie ke 1303 ke Jade Mirror of the Four Unknowns mein. Europe mein Michael Stifel ne 1544 ki apni Arithmetica Integra mein coefficients chhape.

Blaise Pascal (1623–1662) ne arithmetical triangle ko ek alag subject bana diya. Unhone iski kai properties prove ki, iske liye induction ka ek shuruaati roop use kiya, aur ise chance ke problems par lagaya. Unka Traité du triangle arithmétique 1665 mein publish hua, unki death ke teen saal baad, aur zyadatar textbooks triangle ko unhi ke naam se bulati hain.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 7.1 Q1	Practice 7.1 · Q1
NCERT Exercise 7.1 Q2	Practice 7.1 · Q2
NCERT Exercise 7.1 Q3	Practice 7.1 · Q3
NCERT Exercise 7.1 Q4	Practice 7.1 · Q4
NCERT Exercise 7.1 Q5	Practice 7.1 · Q5
NCERT Exercise 7.1 Q6	Practice 7.1 · Q6
NCERT Exercise 7.1 Q7	Practice 7.1 · Q7
NCERT Exercise 7.1 Q8	Practice 7.1 · Q8
NCERT Exercise 7.1 Q9	Practice 7.1 · Q9
NCERT Exercise 7.1 Q10	Practice 7.1 · Q10
NCERT Exercise 7.1 Q11	Practice 7.1 · Q11
NCERT Exercise 7.1 Q12	Practice 7.1 · Q12
NCERT Exercise 7.1 Q13	Practice 7.1 · Q13
NCERT Exercise 7.1 Q14	Practice 7.1 · Q14
NCERT Miscellaneous Exercise Q1	Chapter 7 par miscellaneous exercises · Q1
NCERT Miscellaneous Exercise Q2	Chapter 7 par miscellaneous exercises · Q2
NCERT Miscellaneous Exercise Q3	Chapter 7 par miscellaneous exercises · Q3
NCERT Miscellaneous Exercise Q4	Chapter 7 par miscellaneous exercises · Q4
NCERT Miscellaneous Exercise Q5	Chapter 7 par miscellaneous exercises · Q5
NCERT Miscellaneous Exercise Q6	Chapter 7 par miscellaneous exercises · Q6

Quick check: MCQs

7.1 Introduction2 questions
Pascal's triangle4 questions
7.2.1 Kisi bhi positive integer $n$ ke liye binomial theorem4 questions
7.2.2 Kuch special cases3 questions
Theorem ka use4 questions

Glossary · 10

Key terms (English + Hindi)

binomialद्विपदdvipad: Theek do terms wala algebraic expression, jaise $a + b$ ya $3 x - 2$ .Chapter mein pehli baar yahan aaya →
binomial coefficientद्विपद गुणांकdvipad gunaank: Numbers $^{n} C_{r}$ mein se ek, jo $(a + b)^{n}$ ke terms ko multiply karte hain.
binomial theoremद्विपद प्रमेयdvipad pramey: Har positive integer $n$ ke liye rule $(a + b)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}$ .Chapter mein pehli baar yahan aaya →

combinationसंचयsanchay: Objects ka aisa selection jismein order matter nahi karta; $n$ mein se $r$ objects $^{n} C_{r}$ tareekon se chune ja sakte hain.Chapter mein pehli baar yahan aaya →

factorialक्रमगुणितkramagunit: Positive integer $n$ ke liye $n! = 1 \times 2 \times \dots \times n$ , aur $0! = 1$ .Chapter mein pehli baar yahan aaya →

indexघातांकghaataank: Woh power jis tak expression ko raise kiya jaata hai; $(a + b)^{n}$ mein index $n$ hai.Chapter mein pehli baar yahan aaya →

mathematical inductionगणितीय आगमनganitiya aagaman: Kisi statement ko har positive integer $n$ ke liye prove karne ka tareeka: use $n = 1$ ke liye prove karo, phir prove karo ki $n = k$ ke liye sahi hona $n = k + 1$ ke liye bhi sahi hona zaroori bana deta hai.Chapter mein pehli baar yahan aaya →

Pascal's rule: $1 \leq r \leq n$ ke liye identity $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ .Chapter mein pehli baar yahan aaya →
Pascal's triangleपास्कल त्रिभुजpaaskal tribhuj: Ek triangular array jismein har row ke dono ends par 1 hota hai aur andar ki har entry upar wali do entries ka sum hoti hai; row $n$ mein $^{n} C_{0}, \dots,^{n} C_{n}$ hote hain.Chapter mein pehli baar yahan aaya →

sigma notationसिग्मा संकेतनsigma sanketan: Sum likhne ka chhota tareeka: $\sum_{r = 0}^{n} f (r)$ ka matlab hai $f (0) + f (1) + \dots + f (n)$ .Chapter mein pehli baar yahan aaya →

Binomial Theorem: Class 11 Maths Chapter 7 Hinglish mein

Is chapter mein aap kya seekhenge

7.1 Introduction

7.2 Positive Integral Indices ke liye Binomial Theorem

Pascal's triangle

7.2.1 Kisi bhi positive integer $n$ ke liye binomial theorem

7.2.2 Kuch special cases

Theorem ka use

Practice 7.1: answers ke saath

Chapter 7 par miscellaneous exercises: answers ke saath

Practice questions answers ke saath (NCERT “Entrance jaisi practice”)

Summary aur key points

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye

7.1 Introduction

7.2 Positive Integral Indices ke liye Binomial Theorem

Pascal's triangle

7.2.1 Kisi bhi positive integer n ke liye binomial theorem

7.2.2 Kuch special cases

Theorem ka use

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye

7.2.1 Kisi bhi positive integer $n$ ke liye binomial theorem