Binomial Theorem: Class 11 Maths · Chapter 7

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

55 min
28 practice questions
17 MCQs

What you will learn in Binomial Theorem

Recognise the patterns in the expansions of (a + b)ⁿ for small n and build rows of Pascal's triangle by adding neighbours.
Explain why the coefficients of (a + b)ⁿ are the combinations ⁿCᵣ, and state the binomial theorem in full and in sigma notation.
Prove the binomial theorem by mathematical induction using Pascal's rule ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ.
Expand (x − y)ⁿ, (1 + x)ⁿ and (1 − x)ⁿ, and deduce that the binomial coefficients add up to 2ⁿ and that their alternating sum is 0.
Expand binomials whose terms include reciprocal powers, and trinomials by grouping them as binomials.
Use the theorem to evaluate powers of numbers near 10, 100 or 1000, compare large powers with numbers and approximate powers of numbers near 1.
Prove divisibility and remainder results, such as 100 dividing 11ⁿ − 10n − 1, by expanding (1 + m)ⁿ.

Original companion text aligned to NCERT Mathematics Class 11, Chapter 7, pp. 126–134. Not affiliated with or endorsed by NCERT.

7.1 Introduction

You have used $(a + b)^{2}$ and $(a - b)^{2}$ for years, often as shortcuts: $103^{2} = (100 + 3)^{2} = 10000 + 600 + 9 = 10609$ , with no long multiplication. The cube identities do the same job for third powers. Higher powers are another story. A fixed deposit of ₹10,000 at 3% a year, compounded yearly, is worth $10000 \times (1.03)^{6}$ rupees after six years, and multiplying $1.03$ by itself six times by hand is slow and easy to get wrong.

A binomial (द्विपद) is a sum of two terms, such as $a + b$ , $2 x - 5$ or $1 + 0.03$ . The binomial theorem writes out $(a + b)^{n}$ term by term for any positive integer $n$ , with coefficients you can read off directly. This chapter deals only with positive integer indices. Isaac Newton extended the theorem to negative and fractional indices in the 1660s; that version belongs to later courses.

Begin with a picture of the identity you know best. Fig. 7.1 cuts a square of side $a + b$ into pieces whose areas add up to $(a + b)^{2}$ .

3/3Two of the pieces are equal, so $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .

Fig. 7.1Area model of

(a + b)^{2}

: the two equal

a b

pieces give the middle coefficient 2.

A large square whose top and left sides are each split into a length a and a length b. In the first step the square is whole and its area is written as (a + b) squared. In the second step dashed cuts divide it into four pieces: a square of area a squared, two rectangles of area ab each, and a small square of area b squared, with the sum of the four areas written beside it. In the third step the two equal rectangles are outlined, and the text gives (a + b) squared = a squared + 2ab + b squared, with coefficients 1, 2, 1.

Cutting a cube of edge $a + b$ in the same way gives one $a^{3}$ block, three $a^{2} b$ slabs, three $a b^{2}$ rods and one $b^{3}$ cube, which is why $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ . The coefficients 1, 2, 1 and 1, 3, 3, 1 are the first clues to the pattern this chapter uncovers.

Quick check

1 of 2

Use

(a - b)^{2}

with

a = 100

to find

97^{2}

Show answer

Answer 9409

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

. Forgetting the middle term gives

10000 + 9 = 10009

, which cannot be right since

97^{2} < 100^{2}

7.2 Binomial Theorem for Positive Integral Indices

Multiplying out one power at a time gives the first few expansions (the zeroth power needs $a + b \neq 0$ ). Each line comes from the one above it by one more multiplication by $a + b$ :

\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)

Take the last line. $(a + b)^{4} = (a + b)^{3} (a + b)$ , so we multiply $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ once by $a$ and once by $b$ and collect like terms: $a^{4} + (3 + 1) a^{3} b + (3 + 3) a^{2} b^{2} + (1 + 3) a b^{3} + b^{4}$ . Keep an eye on the sums $3 + 1$ , $3 + 3$ and $1 + 3$ ; they return in Pascal's triangle.

Three patterns hold in every line of (7.1):

The number of terms is one more than the index: $(a + b)^{4}$ has five.
From one term to the next, the power of $a$ drops by 1 and the power of $b$ grows by 1.
Within each term the two powers add up to the index; in $6 a^{2} b^{2}$ , $2 + 2 = 4$ .

Fig. 7.2 makes the patterns visible by drawing each term of $(a + b)^{4}$ as a stack of its four letters.

4/4Every column is 4 cells high: the powers add up to the index.

Fig. 7.2Each term of

(a + b)^{4}

is a product of four letters, so its two powers always add up to 4.NCERT Observations (i)–(iii) · p. 126

Five columns of four square cells stand side by side, with the terms a to the fourth, 4a cubed b, 6a squared b squared, 4a b cubed and b to the fourth written below them. Reading left to right, the columns hold four a cells; three a cells and one b cell; two of each; one a cell and three b cells; and four b cells, with the a cells at the bottom. In the first step the cells are plain. In the second step the a cells are shaded and a row of numbers 4, 3, 2, 1, 0 gives the power of a. In the third step the dashed b cells are shaded and a second row 0, 1, 2, 3, 4 gives the power of b. In the fourth step a double arrow marks every column as 4 cells high, with the note that the powers add up to 4 in every term.

Pascal's triangle

Now strip away the letters and keep only the coefficients, one row per index, each row centred under the one above. The rows for indices 0 to 4 read 1; 1, 1; 1, 2, 1; 1, 3, 3, 1; and 1, 4, 6, 4, 1. We count rows from 0, so the row for index $n$ is called row $n$ .

Two rules generate each row from the one before. Every row begins and ends with 1. Every number inside a row is the sum of the two numbers just above it, to its upper left and upper right; this is the $3 + 1$ , $3 + 3$ , $1 + 3$ bookkeeping from building $(a + b)^{4}$ . Fig. 7.3 grows the triangle with these two rules.

4/4Both edges are all 1s, and every row is symmetric about the middle.

Fig. 7.3Each row starts and ends with 1, and every inside number is the sum of the two above it.NCERT Fig. 7.1, Fig. 7.2 · p. 127

A triangle of hexagonal cells, one row per index, from row 0 at the top to row 6 at the bottom. In the first step only rows 0 to 2 are shown: 1; 1, 1; 1, 2, 1. In the second step rows 3 and 4 appear; the two 3s in row 3 are shaded, arrows run from them into the 6 below, and notes say 3 + 3 = 6 and that each inside number is the sum of the two above. In the third step rows 5 and 6 are added: 1, 5, 10, 10, 5, 1 and 1, 6, 15, 20, 15, 6, 1. In the fourth step a dashed vertical axis runs down the middle, the cells along both slanting edges are shaded and all hold 1, and notes point out that every row reads the same from either end.

Blaise Pascal (1623–1662), a French mathematician, proved so many facts about this array that it now carries his name: Pascal's triangle. Centuries before him, Indian scholars drew it as the Meru-prastāra (मेरु प्रस्तार) while counting the patterns of poetic metres described by Pingala; the note at the end of the chapter tells the story.

ExampleNCERT p. 127

Expanding with a row of the triangle

Use Pascal's triangle to expand $(2 x + y)^{6}$ .

Try it yourself first. Then reveal the solution one step at a time.

The method has a weakness. To expand $(2 x + y)^{15}$ you would first need all the rows up to row 15, and one slip in an early row spoils every row below it. What we want is a formula for row $n$ that does not depend on the rows above it.

Combinations from Chapter 6 supply it. For a non-negative integer $n$ and $0 \leq r \leq n$ , with factorials $n! = 1 \times 2 \times \dots \times n$ and $0! = 1$ ,

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

(7.2)

where

$^{n} C_{r}$: the number of ways to choose $r$ objects from $n$

Test it on row 4: $^{4} C_{0},^{4} C_{1},^{4} C_{2},^{4} C_{3},^{4} C_{4}$ are $1, 4, 6, 4, 1$ , exactly the row. Row 8, computed straight from the formula without rows 5 to 7, is 1, 8, 28, 56, 70, 56, 28, 8, 1. In general, row $n$ of Pascal's triangle is $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ .

Why should a counting formula produce these coefficients? Fig. 7.4 answers the question for $(a + b)^{3}$ .

3/3There are $^{3} C_{r}$ ways to choose which $r$ brackets give $b$ : 1, 3, 3, 1.

Fig. 7.4Choosing

b

from

r

of the 3 brackets can be done in

^{3} C_{r}

ways, so

^{3} C_{r}

is the coefficient of

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

.NCERT Fig. 7.3 · p. 128

At the top is the product (a + b)(a + b)(a + b). In the first step the eight ways of choosing one letter from each bracket are listed in a row: aaa, aab, aba, baa, abb, bab, bba and bbb, with the note 2 × 2 × 2 = 8 choices; every b is printed in bold. In the second step the words slide into four columns headed a cubed, a squared b, a b squared and b cubed according to how many b letters they contain, with 1 way, 3 ways, 3 ways and 1 way written below. In the third step the counts are labelled with the combinations 3C0, 3C1, 3C2 and 3C3.

The same reasoning works for any index. Multiplying out $(a + b)^{n}$ means choosing one letter from each of the $n$ brackets in every possible way and adding the products. A product equals $a^{n - r} b^{r}$ exactly when $b$ came from $r$ of the brackets, and those $r$ brackets can be picked in $^{n} C_{r}$ ways. So $a^{n - r} b^{r}$ turns up $^{n} C_{r}$ times, and its coefficient is $^{n} C_{r}$ . Row 6, for instance, written as combinations:

\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 Binomial theorem for any positive integer $n$

The choosing argument works for every index, and it leads to the general statement:

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)

The counting argument is already a convincing reason, but the standard proof, the one examiners expect, uses mathematical induction. Induction no longer has a chapter of its own in this book, so here is the idea in brief.

3/3Together the two steps give $P (n)$ for every positive integer $n$ .

Fig. 7.5The first domino falls, and each falling domino knocks over the next one, so they all fall.

Six dominoes, numbered 1 to 6, stand in a row on a line, with dots after them to show that the row goes on. In the first step an arrow pushes the first domino, which is shaded, with the note: first, n = 1 is true. In the second step dominoes 1 and 2 have tipped over and the next one is shaded, with the note that case k knocks over case k + 1. In the third step every domino has fallen, with the note: so every case falls.

The proof is shortest in sigma notation. The symbol $\sum_{r = 0}^{n}$ means: write the expression for $r = 0, 1, 2, \dots, n$ in turn and add the results. Reading $a^{0}$ and $b^{0}$ as 1, the theorem becomes

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

(7.5)

The proof also needs one fact from Chapter 6, Pascal's rule, which is exactly the add-the-two-above rule of the triangle:

Pascal's rule

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

(7.6)

where

$r$: any whole number with $1 \leq r \leq n$

TheoremBinomial theorem

For every positive integer $n$ and any numbers $a$ and $b$ , the expansion (7.4) holds; in sigma form, $(a + b)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}$ .

Proof. Let $P (n)$ be the statement (7.5). We check the first case and then show that each case carries the next.

First case. For $n = 1$ the right side is $^{1} C_{0} a +^{1} C_{1} b = a + b$ , which is $(a + b)^{1}$ . So $P (1)$ is true.

From $k$ to $k + 1$ . Fix one positive integer $k$ and assume that $P (k)$ holds:

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

Multiply both sides by $a + b$ . Multiplying by $a$ raises every power of $a$ by one; multiplying by $b$ raises every power of $b$ by one:

\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}

In the second sum put $s = r + 1$ . Then $s$ runs from 1 to $k + 1$ and the general term is $^{k} C_{s - 1} a^{k + 1 - s} b^{s}$ . Renaming $s$ as $r$ , both sums now list the same powers $a^{k + 1 - r} b^{r}$ .

Take the $r = 0$ term out of the first sum and the $r = k + 1$ term out of the second, then pair what is left:

\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}

where $c_{r} =^{k} C_{r} +^{k} C_{r - 1}$ . By Pascal's rule (7.6), $c_{r} =^{k + 1} C_{r}$ . Also $a^{k + 1} =^{k + 1} C_{0} a^{k + 1}$ and $b^{k + 1} =^{k + 1} C_{k + 1} b^{k + 1}$ , since both coefficients are 1. Hence

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

which is exactly $P (k + 1)$ . The first case holds and each case forces the next, so by the principle of mathematical induction $P (n)$ holds for $n = 1, 2, 3, \dots$ without exception.

Fig. 7.6 shows the heart of the proof for $k = 4$ with plain numbers. The coefficients of $a (a + b)^{4}$ and of $b (a + b)^{4}$ are the same row, shifted by one place, and adding the columns produces row 5.

3/3Each column sum is $^{4} C_{r} +^{4} C_{r - 1} =^{5} C_{r}$ , which gives row 5.

Fig. 7.6Multiplying by

a + b

adds row 4 to a shifted copy of itself, giving row 5: Pascal's rule at work.NCERT p. 129

Six columns are headed by the powers a to the fifth, a to the fourth b, a cubed b squared, a squared b cubed, a b to the fourth and b to the fifth. In the first step the row 1, 4, 6, 4, 1 fills the first five columns, labelled times a. In the second step a second copy of the row, labelled times b, slides in one column further right, filling columns two to six. In the third step a line is drawn under both rows and the column sums 1, 5, 10, 10, 5, 1 appear, with the column holding 6 and 4 highlighted and the note 6 + 4 = 10.

ExampleNCERT p. 129

A first expansion with the theorem

Expand $(y + 3)^{5}$ using the binomial theorem.

Try it yourself first. Then reveal the solution one step at a time.

Definition

Binomial coefficients

The numbers $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ that multiply the terms of $(a + b)^{n}$ . Row $n$ of Pascal's triangle lists them in order.

ExampleFor $n = 4$ they are 1, 4, 6, 4, 1.

Reading the theorem closely gives a checklist for any expansion:

Number of terms. $(a + b)^{n}$ has $n + 1$ terms, one for each $r = 0, 1, \dots, n$ .
Powers. From left to right the power of $a$ steps down from $n$ to 0 while the power of $b$ steps up from 0 to $n$ .
Degree. In every term the two powers add up to $n$ , because $(n - r) + r = n$ .
Symmetry. Since $^{n} C_{r} =^{n} C_{n - r}$ , the coefficients read the same from both ends.

InteractiveBinomial lab

Two tabs. In the Triangle tab a slider shows rows 0 to 10 of Pascal's triangle; tapping an entry highlights it and the two entries above it, and a panel gives its row and position, its value from the factorial formula, the sum of the two numbers above it, its mirror-image entry and the sum of its row as a power of 2. In the Expand tab a slider sets the index n from 1 to 10. In Letters mode, buttons set whole numbers p and q from −5 to 5 and a table lists, for each r, the binomial coefficient, the product of powers and the simplified term of (px + qy) to the n, followed by the full expansion and a check that the coefficients add up to (p + q) to the n. In Numbers mode, A is 10, 100 or 1000 and B is a whole number from −9 to 9; the table lists each term of (A + B) to the n with a running total that ends at the exact value.

Fig. 7.7Tap any entry of Pascal's triangle to see it as

^{n} C_{r}

and as the sum of the two above, or expand

(p x + q y)^{n}

and powers such as

(100 - 3)^{n}

term by term.

Quick check

1 of 4

How many terms does the expansion of

(3 p - 2 q)^{11}

have?

Show answer

Answer 12

One more than the index:

11 + 1 = 12

. Answering 11 forgets the term with

r = 0

7.2.2 Some special cases

The theorem holds for any numbers $a$ and $b$ , so we may feed it special values. Three choices are used again and again.

(i) A difference. Put $a = x$ and $b = - y$ . Since $(- y)^{r}$ is $y^{r}$ for even $r$ and $- y^{r}$ for odd $r$ , every term keeps its size but the signs alternate:

\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

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

Try it yourself first. Then reveal the solution one step at a time.

(ii) One plus something. Put $a = 1$ and $b = x$ . Every power of 1 is 1, so only the powers of $x$ remain:

\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)

This form does most of the work in approximations and divisibility proofs, as the examples below show. Putting $x = 1$ in it adds up a whole row of the triangle:

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

(7.9)

There is a counting reason too. $^{n} C_{r}$ counts the subsets with $r$ elements of a set with $n$ elements, so the left side counts all its subsets. Building a subset means deciding in or out for each element, which gives $2 \times 2 \times \dots \times 2 = 2^{n}$ choices.

(iii) One minus something. Put $a = 1$ and $b = - x$ :

\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)

With $x = 1$ the left side is $0^{n} = 0$ for every positive integer $n$ , so the coefficients with alternating signs cancel completely:

\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 checks both facts on the first rows. Only row 0 breaks the second one, because $(1 - 1)^{0} = 1$ ; that row is excluded, since $n$ is a positive integer.

3/3With signs $+, -, +, -, \dots$ each total is 0, except row 0.

Fig. 7.8Row

n

adds up to

2^{n}

; with alternating signs, every row after row 0 adds up to 0.NCERT p. 131

Rows 0 to 5 of Pascal's triangle are written as a left-aligned staircase, one row per line, from 1 at the top to 1, 5, 10, 10, 5, 1 at the bottom. In the first step only the rows are shown. In the second step a shaded column headed sum gives the total of each row: 1, 2, 4, 8, 16 and 32, doubling each time. In the third step minus signs appear in front of every second number in each row, and a second column headed plus minus plus minus gives the totals with these alternating signs: 1 for the top row and 0 for every other row.

Adding (7.9) and (7.11) cancels the coefficients in odd positions, and subtracting them cancels those in even positions. So each half of a row carries exactly half of the total:

\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

1 of 3

The coefficient of

x^{3}

(2 - x)^{7}

560

- 560

- 280

- 35

Show answer

Answer (B) $- 560$

The term is

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

. The sign comes from the odd power of

- x

;

- 280

uses

2^{3}

instead of

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

Using the theorem

Example 7.1NCERT Example 1 · p. 131

A binomial with a reciprocal power

Expand ${(3 x - \frac{1}{x^{2}})}^{4}$ , where $x \neq 0$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 7.2NCERT Example 2 · p. 131

A fourth power without long multiplication

Find $97^{4}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 7.3NCERT Example 3 · p. 131

Comparing a huge power with a number

Which is larger, $(1.002)^{5000}$ or 10?

Try it yourself first. Then reveal the solution one step at a time.

Why is the two-term bound safe, and how quickly does it lose accuracy? Fig. 7.9 compares $(1.1)^{n}$ with its first two and first three terms.

3/3Adding $^{n} C_{2} (0.1)^{2}$ gets closer, but is still below.

Fig. 7.9All the dropped terms are positive, so each truncation lies below

(1.1)^{n}

, and the gap grows with

n

.NCERT p. 132

A graph with n from 0 to 20 across and values from 0 to 7 up. In the first step dots show 1.1 to the power n at every whole number n, rising from 1 to about 6.7 and curving upward. In the second step a dashed straight line for the first two terms, 1 + 0.1n, is added; it starts at the same point but stays below the dots and reaches only 3 at n = 20, and a double arrow at n = 16 marks the gap, labelled gap greater than 0. In the third step a dotted curve for the first three terms is added; it lies between the line and the dots, close to the dots for small n, and reaches 4.9 at n = 20.

Example 7.4NCERT Example 4 · p. 132

The last two digits of a power of 11

Show that $11^{n} - 10 n - 1$ is divisible by 100 for every positive integer $n$ . Hence find the last two digits of $11^{7}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example

Back to the fixed deposit

Estimate $10000 \times (1.03)^{6}$ , the deposit from Section 7.1, using the first three terms of the expansion. How good is the estimate?

Try it yourself first. Then reveal the solution one step at a time.

Example

Three terms treated as two

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

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Using the first three terms,

(1.01)^{8}

is approximately

1.0800

1.0882

1.0280

1.0828

Show answer

Answer (D) $1.0828$

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

; the exact value is 1.08286 to five places. 1.0800 stops after two terms.

Practice · 14 questions

Practice 7.1: questions with answers

Short answerEasyNCERT Exercise 7.1 Q1
Expand $(1 - 3 x)^{5}$ .
Show solution
Answer $1 - 15 x + 90 x^{2} - 270 x^{3} + 405 x^{4} - 243 x^{5}$
Use (7.10) with $3 x$ in place of $x$ . The term with $r$ factors of $- 3 x$ is $^{5} C_{r} (- 3 x)^{r}$ :
$\begin{aligned} (1 - 3 x)^{5} \\ = 1 - 5 (3 x) + 10 (9 x^{2}) \\ - 10 (27 x^{3}) + 5 (81 x^{4}) \\ - 243 x^{5} \end{aligned}$
So $(1 - 3 x)^{5} = 1 - 15 x + 90 x^{2} - 270 x^{3} + 405 x^{4} - 243 x^{5}$ . Check with $x = 1$ : the sum is $- 32 = (- 2)^{5}$ .
Short answerMediumNCERT Exercise 7.1 Q2
Expand ${(\frac{x}{3} + \frac{3}{x})}^{4}$ , $x \neq 0$ .
Show solution
Answer $\frac{x^{4}}{81} + \frac{4 x^{2}}{9} + 6 + \frac{36}{x^{2}} + \frac{81}{x^{4}}$
With $a = \frac{x}{3}$ , $b = \frac{3}{x}$ and row 4 = 1, 4, 6, 4, 1:
$\begin{aligned} r = 0 & : \frac{x^{4}}{81} \\ r = 1 & : 4 \cdot \frac{x^{3}}{27} \cdot \frac{3}{x} = \frac{4 x^{2}}{9} \\ r = 2 & : 6 \cdot \frac{x^{2}}{9} \cdot \frac{9}{x^{2}} = 6 \\ r = 3 & : 4 \cdot \frac{x}{3} \cdot \frac{27}{x^{3}} = \frac{36}{x^{2}} \\ r = 4 & : \frac{81}{x^{4}} \end{aligned}$
Add the five terms. The middle term is free of $x$ because there $x$ and $\frac{1}{x}$ appear equally often.
Short answerEasyNCERT Exercise 7.1 Q3
Expand $(2 x - 5)^{4}$ .
Show solution
Answer $16 x^{4} - 160 x^{3} + 600 x^{2} - 1000 x + 625$
$\begin{aligned} (2 x - 5)^{4} \\ = (2 x)^{4} - 4 (2 x)^{3} (5) \\ + 6 (2 x)^{2} (5^{2}) \\ - 4 (2 x) (5^{3}) + 5^{4} \end{aligned}$
$= 16 x^{4} - 160 x^{3} + 600 x^{2} - 1000 x + 625$ . Check with $x = 1$ : $16 - 160 + 600 - 1000 + 625 = 81 = (- 3)^{4}$ .
Short answerMediumNCERT Exercise 7.1 Q4
Expand ${(\frac{x}{2} - \frac{1}{x})}^{6}$ , $x \neq 0$ .
Show solution
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 is 1, 6, 15, 20, 15, 6, 1 and the powers of $\frac{1}{2}$ from ${(\frac{1}{2})}^{6}$ down to ${(\frac{1}{2})}^{0}$ are $\frac{1}{64}, \frac{1}{32}, \frac{1}{16}, \frac{1}{8}, \frac{1}{4}, \frac{1}{2}, 1$ . Multiplying and attaching the alternating signs:
$\begin{aligned} \frac{1}{64}, - \frac{6}{32}, \frac{15}{16}, - \frac{20}{8}, \\ \frac{15}{4}, - \frac{6}{2}, 1 \end{aligned}$
These multiply $x^{6}, x^{4}, x^{2}, x^{0}, x^{- 2}, x^{- 4}, x^{- 6}$ . Simplifying, $\frac{6}{32} = \frac{3}{16}$ , $\frac{20}{8} = \frac{5}{2}$ and $\frac{6}{2} = 3$ , which gives the answer.
Short answerEasyNCERT Exercise 7.1 Q5
Expand ${(x - \frac{1}{x})}^{5}$ , $x \neq 0$ .
Show solution
Answer $x^{5} - 5 x^{3} + 10 x - \frac{10}{x} + \frac{5}{x^{3}} - \frac{1}{x^{5}}$
The term with $r$ factors of $- \frac{1}{x}$ is $^{5} C_{r} (- 1)^{r} x^{5 - 2 r}$ . With row 5 = 1, 5, 10, 10, 5, 1 and alternating signs, the expansion is $x^{5} - 5 x^{3} + 10 x - 10 x^{- 1} + 5 x^{- 3} - x^{- 5}$ .
NumericalEasyNCERT Exercise 7.1 Q6
Use the binomial theorem to evaluate $95^{3}$ .
Show solution
$\begin{aligned} (100 - 5)^{3} \\ = 1 000 000 - 150 000 \\ + 7 500 - 125 \end{aligned}$
$= 857 375$ .
NumericalEasyNCERT Exercise 7.1 Q7
Use the binomial theorem to evaluate $103^{4}$ .
Show solution
$\begin{aligned} (100 + 3)^{4} \\ = 100 000 000 + 12 000 000 \\ + 540 000 + 10 800 + 81 \end{aligned}$
$= 112 550 881$ .
NumericalEasyNCERT Exercise 7.1 Q8
Use the binomial theorem to evaluate $1001^{3}$ .
Show solution
$\begin{aligned} (1000 + 1)^{3} & = 10^{9} + 3 \times 10^{6} \\ + 3 \times 10^{3} + 1 \end{aligned}$
$= 1 003 003 001$ .
NumericalEasyNCERT Exercise 7.1 Q9
Use the binomial theorem to evaluate $999^{4}$ .
Show solution
$\begin{aligned} (1000 - 1)^{4} \\ = 10^{12} - 4 \times 10^{9} \\ + 6 \times 10^{6} - 4 \times 10^{3} + 1 \end{aligned}$
$= 996 005 996 001$ .
Quick answerMediumNCERT Exercise 7.1 Q10
Without a calculator, decide which is larger: $(1.05)^{400}$ or 20.
20 $(1.05)^{400}$
Show solution
$(1.05)^{400} = 1 + 400 (0.05) + (positive terms) > 1 + 20 = 21$ , so $(1.05)^{400} > 20$ . (Its actual size is about $3 \times 10^{8}$ .)
11
Short answerMediumNCERT Exercise 7.1 Q11
Differences of fifth powers.
1. Expand and simplify $(x + y)^{5} - (x - y)^{5}$ .
  Show solution
  Answer $2 (5 x^{4} y + 10 x^{2} y^{3} + y^{5})$
  In $(x - y)^{5}$ the terms with odd powers of $y$ change sign, so in the difference the even-power terms cancel and the odd-power terms double: $2 (^{5} C_{1} x^{4} y +^{5} C_{3} x^{2} y^{3} +^{5} C_{5} y^{5}) = 10 x^{4} y + 20 x^{2} y^{3} + 2 y^{5}$ .
2. Hence evaluate $(2 + \sqrt{3})^{5} - (2 - \sqrt{3})^{5}$ .
  Show solution
  Answer $418 \sqrt{3}$
  Put $x = 2$ and $y = \sqrt{3}$ , so $y^{3} = 3 \sqrt{3}$ and $y^{5} = 9 \sqrt{3}$ :
  $\begin{aligned} 2 (80 \sqrt{3} + 120 \sqrt{3} + 9 \sqrt{3}) \\ = 418 \sqrt{3} \end{aligned}$
12
Short answerMediumNCERT Exercise 7.1 Q12
Sums of fourth powers.
1. Expand and simplify $(x + 1)^{4} + (x - 1)^{4}$ .
  Show solution
  Answer $2 x^{4} + 12 x^{2} + 2$
  The odd-power terms $4 x^{3}$ and $4 x$ cancel, and the rest double: $2 (x^{4} + 6 x^{2} + 1) = 2 x^{4} + 12 x^{2} + 2$ .
2. Hence evaluate $(\sqrt{5} + 1)^{4} + (\sqrt{5} - 1)^{4}$ .
  Show solution
  With $x = \sqrt{5}$ : $x^{2} = 5$ and $x^{4} = 25$ , so the value is $2 (25 + 30 + 1) = 112$ .
ProofHardNCERT Exercise 7.1 Q13
Show that $5^{n + 1} - 4 n - 5$ is divisible by 16 for every positive integer $n$ .
Show solution
By (7.8) with $x = 4$ :
$\begin{aligned} 5^{n + 1} & = 5 (1 + 4)^{n} \\ = 5 (1 + 4 n +^{n} C_{2} 4^{2} \\ + \dots + 4^{n}) \\ = 5 + 20 n + 80 k \end{aligned}$
where $k =^{n} C_{2} +^{n} C_{3} 4 + \dots + 4^{n - 2}$ is an integer ( $k = 0$ when $n = 1$ ). Then $5^{n + 1} - 4 n - 5 = 16 n + 80 k = 16 (n + 5 k)$ , a multiple of 16. Check: $n = 2$ gives $125 - 8 - 5 = 112 = 16 \times 7$ .
ProofMediumNCERT Exercise 7.1 Q14
Prove that $\sum_{r = 0}^{n} 2^{r}^{n} C_{r} = 3^{n}$ .
Show solution
Put $x = 2$ in $(1 + x)^{n} = \sum_{r = 0}^{n}^{n} C_{r} x^{r}$ : the left side is $3^{n}$ and the right side is $\sum 2^{r}^{n} C_{r}$ . For $n = 3$ : $1 + 6 + 12 + 8 = 27 = 3^{3}$ .

Practice · 6 questions

Miscellaneous exercises on Chapter 7: questions with answers

1
ProofHardNCERT Miscellaneous Exercise Q1
A divisibility pattern.
1. Show that $17^{n} - 9^{n}$ is divisible by 8 for every positive integer $n$ .
  Show solution
  $(9 + 8)^{n} = 9^{n} +^{n} C_{1} 9^{n - 1} \cdot 8 +^{n} C_{2} 9^{n - 2} \cdot 8^{2} + \dots + 8^{n}$ . Every term after $9^{n}$ contains the factor 8, so $17^{n} - 9^{n} = 8 (^{n} C_{1} 9^{n - 1} +^{n} C_{2} 9^{n - 2} \cdot 8 + \dots + 8^{n - 1})$ , an integer multiple of 8.
2. Explain why the same idea shows that $a - b$ divides $a^{n} - b^{n}$ for any distinct integers $a$ and $b$ .
  Show solution
  Write $a = b + (a - b)$ and expand $(b + (a - b))^{n} = b^{n} +^{n} C_{1} b^{n - 1} (a - b) + \dots + (a - b)^{n}$ . Every term after $b^{n}$ contains $a - b$ , so $a^{n} - b^{n} = (a - b) \times (an integer)$ .
NumericalMediumNCERT Miscellaneous Exercise Q2
Write $(\sqrt{3} + 1)^{6} - (\sqrt{3} - 1)^{6}$ in the form $k \sqrt{3}$ . Enter $k$ .
Show solution
$(x + y)^{6} - (x - y)^{6} = 2 (6 x^{5} y + 20 x^{3} y^{3} + 6 x y^{5})$ . With $x = \sqrt{3}$ , $y = 1$ : $x^{5} = 9 \sqrt{3}$ and $x^{3} = 3 \sqrt{3}$ , so the value is $2 (54 \sqrt{3} + 60 \sqrt{3} + 6 \sqrt{3}) = 240 \sqrt{3}$ . So $k = 240$ .
Short answerHardNCERT Miscellaneous Exercise Q3
Simplify ${(x + \sqrt{x^{2} - 4})}^{4} + {(x - \sqrt{x^{2} - 4})}^{4}$ , where $x^{2} \geq 4$ .
Show solution
Answer $16 x^{4} - 64 x^{2} + 32$
$\begin{aligned} (x + s)^{4} + (x - s)^{4} \\ = 2 (x^{4} + 6 x^{2} s^{2} + s^{4}) \end{aligned}$
With $s^{2} = x^{2} - 4$ and $s^{4} = x^{4} - 8 x^{2} + 16$ :
$\begin{aligned} 2 (x^{4} + 6 x^{4} - 24 x^{2} \\ + x^{4} - 8 x^{2} + 16) \\ = 16 x^{4} - 64 x^{2} + 32 \end{aligned}$
Check with $x = 3$ : $s = \sqrt{5}$ and $(3 + \sqrt{5})^{4} + (3 - \sqrt{5})^{4} = 752 = 16 (81) - 64 (9) + 32$ .
NumericalMediumNCERT Miscellaneous Exercise Q4
Use the first three terms of the expansion to approximate $(0.98)^{6}$ , correct to three decimal places.
Show solution
$(1 - 0.02)^{6} \approx 1 - 6 (0.02) + 15 (0.02)^{2} = 1 - 0.12 + 0.006 = 0.886$ . The next term, $- 20 (0.02)^{3} = - 0.00016$ , does not change the third decimal; the exact value is $0.88584 \dots$
Short answerHardNCERT Miscellaneous Exercise Q5
Expand ${(1 + x - \frac{1}{x})}^{3}$ , $x \neq 0$ .
Show solution
Answer $x^{3} + 3 x^{2} - 5 + \frac{3}{x^{2}} - \frac{1}{x^{3}}$
$(1 + u)^{3} = 1 + 3 u + 3 u^{2} + u^{3}$ , with
$\begin{aligned} u^{2} & = x^{2} - 2 + \frac{1}{x^{2}} \\ u^{3} & = x^{3} - 3 x + \frac{3}{x} - \frac{1}{x^{3}} \end{aligned}$
Adding $1 + (3 x - \frac{3}{x}) + (3 x^{2} - 6 + \frac{3}{x^{2}}) + u^{3}$ , the $3 x$ and $\frac{3}{x}$ terms cancel, leaving $x^{3} + 3 x^{2} - 5 + \frac{3}{x^{2}} - \frac{1}{x^{3}}$ . Check with $x = 1$ : $1 + 3 - 5 + 3 - 1 = 1 = 1^{3}$ .
Short answerHardNCERT Miscellaneous Exercise Q6
Expand $(x^{2} - 2 x + 2)^{3}$ by writing it as a binomial.
Show solution
Answer $x^{6} - 6 x^{5} + 18 x^{4} - 32 x^{3} + 36 x^{2} - 24 x + 8$
$(v + 2)^{3} = v^{3} + 6 v^{2} + 12 v + 8$ , with $v^{2} = x^{4} - 4 x^{3} + 4 x^{2}$ and $v^{3} = x^{6} - 6 x^{5} + 12 x^{4} - 8 x^{3}$ :
$\begin{aligned} x^{6} - 6 x^{5} + 12 x^{4} - 8 x^{3} \\ + 6 x^{4} - 24 x^{3} + 24 x^{2} \\ + 12 x^{2} - 24 x + 8 \end{aligned}$
Collecting: $x^{6} - 6 x^{5} + 18 x^{4} - 32 x^{3} + 36 x^{2} - 24 x + 8$ . Check with $x = 1$ : the sum is $1 = 1^{3}$ .

Recap

Binomial Theorem: key points and summary

Each power of a binomial comes from the one before: $(a + b)^{n + 1} = (a + b)^{n} (a + b)$ (Section 7.2).
$(a + b)^{n}$ has $n + 1$ terms; the power of $a$ falls from $n$ to 0 as the power of $b$ rises from 0 to $n$ , and in every term the two powers add up to $n$ (three patterns).
Pascal's triangle, the Meru-prastāra, lists the coefficients: 1 at both ends of each row, and every inside entry is the sum of the two above it (Pascal's triangle).
Row $n$ is $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ ((7.2)): the coefficient of $a^{n - r} b^{r}$ counts the ways to choose the $r$ brackets that supply $b$ .
Binomial theorem: $(a + b)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}$ for every positive integer $n$ (Theorem), proved by induction with Pascal's rule ((7.6)).
$(x - y)^{n}$ has the same coefficients with alternating signs ((7.7)).
$(1 + x)^{n} = \sum^{n} C_{r} x^{r}$ ; at $x = 1$ the binomial coefficients add up to $2^{n}$ , and with alternating signs to 0 ((7.9), (7.11)).
Powers of numbers near 10, 100 or 1000 are quick to evaluate after writing the number as a round number plus or minus a small one (Example 7.2).
For $x > 0$ every term of $(1 + x)^{n}$ is positive, so the first few terms give a lower bound or a close estimate (Example 7.3, the deposit estimate).
Writing a base as $1 + m$ and expanding proves divisibility facts such as $100 ∣ 11^{n} - 10 n - 1$ (Example 7.4).

Flash cards · 6

Formulas at a glance

1 / 6

From historyNCERT Historical Note · p. 134

From Meru-prastāra to Pascal

Indian prosody gave these numbers one of their earliest homes. Pingala's Chhandaḥśāstra, a work on poetic metre composed more than two thousand years ago, asks how many patterns of long and short syllables a line can have, and how many of them contain a given number of short syllables. The answers are binomial coefficients. Around the 10th century the commentator Halāyudha explained Pingala's rules with a triangular figure, the Meru-prastāra ('the staircase of Mount Meru'), built by adding neighbouring numbers.

The same triangle appears around 1000 CE in the work of al-Karajī in Baghdad, and in China in Yang Hui's book of 1261 and in Zhu Shijie's Jade Mirror of the Four Unknowns of 1303. In Europe, Michael Stifel printed the coefficients in his Arithmetica Integra of 1544.

Blaise Pascal (1623–1662) made the arithmetical triangle a subject of its own. He proved many of its properties, used an early form of induction to do so, and applied it to problems of chance. His Traité du triangle arithmétique was published in 1665, three years after his death, and most textbooks name the triangle after him.

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In your NCERT textbook	Practice question here
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	Miscellaneous exercises on Chapter 7 · Q1
NCERT Miscellaneous Exercise Q2	Miscellaneous exercises on Chapter 7 · Q2
NCERT Miscellaneous Exercise Q3	Miscellaneous exercises on Chapter 7 · Q3
NCERT Miscellaneous Exercise Q4	Miscellaneous exercises on Chapter 7 · Q4
NCERT Miscellaneous Exercise Q5	Miscellaneous exercises on Chapter 7 · Q5
NCERT Miscellaneous Exercise Q6	Miscellaneous exercises on Chapter 7 · Q6

Quick check: MCQs on Binomial Theorem

7.1 Introduction2 questions
Pascal's triangle4 questions
7.2.1 Binomial theorem for any positive integer $n$4 questions
7.2.2 Some special cases3 questions
Using the theorem4 questions

Glossary · 10

Key terms (English–Hindi)

binomialद्विपदdvipad: An algebraic expression with exactly two terms, such as $a + b$ or $3 x - 2$ .First seen in the chapter →
binomial coefficientद्विपद गुणांकdvipad gunaank: One of the numbers $^{n} C_{r}$ that multiply the terms of $(a + b)^{n}$ .
binomial theoremद्विपद प्रमेयdvipad pramey: The rule $(a + b)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}$ for every positive integer $n$ .First seen in the chapter →

combinationसंचयsanchay: A selection of objects in which order does not matter; $r$ objects can be selected from $n$ in $^{n} C_{r}$ ways.First seen in the chapter →

factorialक्रमगुणितkramagunit: $n! = 1 \times 2 \times \dots \times n$ for a positive integer $n$ , with $0! = 1$ .First seen in the chapter →

indexघातांकghaataank: The power to which an expression is raised; in $(a + b)^{n}$ the index is $n$ .First seen in the chapter →

mathematical inductionगणितीय आगमनganitiya aagaman: A way to prove a statement for every positive integer $n$ : prove it for $n = 1$ , then prove that its truth for $n = k$ forces its truth for $n = k + 1$ .First seen in the chapter →

Pascal's rule: The identity $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ for $1 \leq r \leq n$ .First seen in the chapter →
Pascal's triangleपास्कल त्रिभुजpaaskal tribhuj: A triangular array with 1 at both ends of each row and every inside entry the sum of the two above it; row $n$ lists $^{n} C_{0}, \dots,^{n} C_{n}$ .First seen in the chapter →

sigma notationसिग्मा संकेतनsigma sanketan: A compact way to write a sum: $\sum_{r = 0}^{n} f (r)$ means $f (0) + f (1) + \dots + f (n)$ .First seen in the chapter →

Binomial Theorem: Class 11 Maths · Chapter 7

What you will learn in Binomial Theorem

7.1 Introduction

7.2 Binomial Theorem for Positive Integral Indices

Pascal's triangle

7.2.1 Binomial theorem for any positive integer $n$

7.2.2 Some special cases

Using the theorem

Practice 7.1: questions with answers

Miscellaneous exercises on Chapter 7: questions with answers

Practice questions with answers (NCERT “Entrance-style practice”)

Binomial Theorem: key points and summary

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Key terms (English–Hindi)

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7.1 Introduction

7.2 Binomial Theorem for Positive Integral Indices

Pascal's triangle

7.2.1 Binomial theorem for any positive integer n

7.2.2 Some special cases

Using the theorem

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7.2.1 Binomial theorem for any positive integer $n$