vMathematics is a most exact science and its conclusions are capable of

absolute proofs. – C.P. STEINMETZv

8.1 Introduction

In earlier classes, we have learnt how to find the squares

and cubes of binomials like a + b and a – b. Using them, we

could evaluate the numerical values of numbers like

(98)

2

= (100 – 2)

2

, (999)

3

= (1000 – 1)

3

, etc. However, for

higher powers like (98)

5

, (101)

6

, etc., the calculations become

difficult by using repeated multiplication. This difficulty was

overcome by a theorem known as binomial theorem. It gives

an easier way to expand (a + b)

n

, where n is an integer or a

rational number. In this Chapter, we study binomial theorem

for positive integral indices only.

8.2 Binomial Theorem for Positive Integral Indices

Let us have a look at the following identities done earlier:

(a+ b)

0

= 1 a + b ≠ 0

(a+ b)

1

= a + b

(a+ b)

2

= a

2

+ 2ab + b

2

(a+ b)

3

= a

3

+ 3a

2

b + 3

ab

2

+ b

3

(a+ b)

4

= (a + b)

3

(a + b) = a

4

+ 4a

3

b + 6a

2

b

2

+ 4ab

3

+ b

4

In these expansions, we observe that

(i) The total number of terms in the expansion is one more than the index. For

example, in the expansion of (a + b)

2

, number of terms is 3 whereas the index of

(a + b)

2

is 2.

(ii) Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the

second quantity ‘b’ increase by 1, in the successive terms.

(iii) In each term of the expansion, the sum of the indices of a and b is the same and

is equal to the index of a + b.

8

Chapter

Blaise Pascal

(1623-1662)

BINOMIAL THEOREM

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BINOMIAL THEOREM 161

We now arrange the coefficients in these expansions as follows (Fig 8.1):

Do we observe any pattern in this table that will help us to write the next row? Yes we

do. It can be seen that the addition of 1’s in the row for index 1 gives rise to 2 in the row

for index 2. The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in

the row for index 3 and so on. Also, 1 is present at the beginning and at the end of each

row. This can be continued till any index of our interest.

We can extend the pattern given in Fig 8.2 by writing a few more rows.

Pascal’s Triangle

The structure given in Fig 8.2 looks like a triangle with 1 at the top vertex and running

down the two slanting sides. This array of numbers is known as Pascal’s triangle,

after the name of French mathematician Blaise Pascal. It is also known as Meru

Prastara

by Pingla.

Expansions for the higher powers of a binomial are also possible by using Pascal’s

triangle. Let us expand (2x

+ 3y)

5

by using Pascal’s triangle.

The row for index 5 is

1

5 10

10 5 1

Using this row and our observations (i), (ii) and (iii), we get

(2x + 3y)

5

= (2x)

5

+ 5(2x)

4

(3y) + 10(2x)

3

(3y)

2

+10 (2x)

2

(3y)

3

+ 5(2x)(3y)

4

+(3y)

5

= 32x

5

+ 240x

4

y + 720x

3

y

2

+ 1080x

2

y

3

+ 810xy

4

+ 243y

5

.

Fig 8.1

Fig 8.2

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162 MATHEMATICS

Now, if we want to find the expansion of (2x + 3y)

12

, we are first required to get

the row for index 12. This can be done by writing all the rows of the Pascal’s triangle

till index 12. This is a slightly lengthy process. The process, as you observe, will become

more difficult, if we need the expansions involving still larger powers.

We thus try to find a rule that will help us to find the expansion of the binomial for

any power without writing all the rows of the Pascal’s triangle, that come before the

row of the desired index.

For this, we make use of the concept of combinations studied earlier to rewrite

the numbers in the Pascal’s triangle. We know that

!

C

!( )!

n

r

n

r n – r

=

, 0 ≤ r ≤ n and

n is a non-negative integer. Also,

n

C

0

= 1 =

n

C

n

The Pascal’s triangle can now be rewritten as (Fig 8.3)

Observing this pattern, we can now write the row of the Pascal’s triangle for any index

without writing the earlier rows. For example, for the index 7 the row would be

7

C

0

7

C

1

7

C

2

7

C

3

7

C

4

7

C

5

7

C

6

7

C

7.

Thus, using this row and the observations (i), (ii) and (iii), we have

(a + b)

7

=

7

C

0

a

7

+ 7C

1

a

6

b +

7

C

2

a

5

b

2

+

7

C

3

a

4

b

3

+ 7C

4

a

3

b

4

+

7

C

5

a

2

b

5

+

7

C

6

ab

6

+

7

C

7

b

7

An expansion of a binomial to any positive integral index say n can now be visualised

using these observations. We are now in a position to write the expansion of a binomial

to any positive integral index.

Fig 8.3 Pascal’s triangle

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BINOMIAL THEOREM 163

8.2.1 Binomial theorem for any positive integer n,

(a + b)

n

=

n

C

0

a

n

+

n

C

1

a

n–1

b +

n

C

2

a

n–2

b

2

+ ...+

n

C

n – 1

a.b

n–1

+

n

C

n

b

n

Proof The proof is obtained by applying principle of mathematical induction.

Let the given statement be

P(n) : (a + b)

n

=

n

C

0

a

n

+

n

C

1

a

n – 1

b +

n

C

2

a

n – 2

b

2

+ ...+

n

C

n–1

a.b

n – 1

+

n

C

n

b

n

For n = 1, we have

P (1) : (

a + b)

1

=

1

C

0

a

1

+

1

C

1

b

1

= a + b

Thus, P (1) is true.

Suppose P (

k) is true for some positive integer k, i.e.

(a + b)

k

=

k

C

0

a

k

+

k

C

1

a

k – 1

b +

k

C

2

a

k – 2

b

2

+ ...+

k

C

k

b

k

... (1)

We shall prove that P(k + 1) is also true, i.e.,

(a + b)

k + 1

=

k + 1

C

0

a

k + 1

+

k + 1

C

1

a

k

b +

k + 1

C

2

a

k – 1

b

2

+ ...+

k + 1

C

k+1

b

k + 1

Now, (a + b)

k + 1

= (a + b) (a + b)

k

= (a + b) (

k

C

0

a

k

+

k

C

1

a

k – 1

b +

k

C

2

a

k – 2

b

2

+...+

k

C

k – 1

ab

k – 1

+

k

C

k

b

k

)

[from (1)]

=

k

C

0

a

k +

1

+

k

C

1

a

k

b +

k

C

2

a

k – 1

b

2

+...+

k

C

k – 1

a

2

b

k – 1

+

k

C

k

ab

k

+

k

C

0

a

k

b

+

k

C

1

a

k

– 1

b

2

+

k

C

2

a

k

– 2

b

3

+...+

k

C

k-1

ab

k

+

k

C

k

b

k + 1

[by actual multiplication]

=

k

C

0

a

k + 1

+ (

k

C

1

+

k

C

0

) a

k

b + (

k

C

2

+

k

C

1

)a

k – 1

b

2

+ ...

+ (

k

C

k

+

k

C

k–1

) ab

k

+

k

C

k

b

k + 1

[grouping like terms]

=

k + 1

C

0

a

k + 1

+

k + 1

C

1

a

k

b +

k + 1

C

2

a

k – 1

b

2

+...+

k + 1

C

k

ab

k

+

k + 1

C

k + 1

b

k +1

(by using

k + 1

C

0

=1,

k

C

r

+

k

C

r–1

=

k + 1

C

r

and

k

C

k

= 1=

k + 1

C

k + 1

)

Thus, it has been proved that P (k + 1) is true whenever P(k) is true. Therefore, by

principle of mathematical induction, P(n) is true for every positive integer n.

We illustrate this theorem by expanding (x + 2)

6

:

(x + 2)

6

=

6

C

0

x

6

+

6

C

1

x

5

.2 +

6

C

2

x

4

2

2

+

6

C

3

x

3

.2

3

+

6

C

4

x

2

.2

4

+

6

C

5

x.2

5

+

6

C

6

.2

6

.

= x

6

+ 12x

5

+ 60x

4

+ 160x

3

+

240x

2

+ 192x + 64

Thus (x + 2)

6

= x

6

+ 12x

5

+ 60x

4

+ 160x

3

+

240x

2

+ 192x + 64.

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164 MATHEMATICS

Observations

1. The notation

∑

=

−

n

k

kkn

k

n

ba

0

C

stands for

n

C

0

a

n

b

0

+

n

C

1

a

n–1

b

1

+ ...+

n

C

r

a

n–r

b

r

+ ...+

n

C

n

a

n–n

b

n

, where b

0

= 1 = a

n–n

.

Hence the theorem can also be stated as

∑

=

−

=+

n

k

kkn

k

nn

baba

0

C)(

.

2. The coefficients

n

C

r

occuring in the binomial theorem are known as binomial

coefficients.

3. There are (n+1) terms in the expansion of (a+b)

n

, i.e., one more than the index.

4. In the successive terms of the expansion the index of a goes on decreasing by

unity. It is n in the first term, (n–1) in the second term, and so on ending with zero

in the last term. At the same time the index of b increases by unity, starting with

zero in the first term, 1 in the second and so on ending with n in the last term.

5. In the expansion of (a+b)

n

, the sum of the indices of a and b is n + 0 = n in the

first term, (n – 1) + 1 = n in the second term and so on 0 + n = n in the last term.

Thus, it can be seen that the sum of the indices of a and b is n in every term of the

expansion.

8.2.2 Some special cases In the expansion of (a + b)

n

,

(i) Taking a = x and b = – y, we obtain

(x

– y)

n

= [x + (–y)]

n

=

n

C

0

x

n

+

n

C

1

x

n – 1

(–y) +

n

C

2

x

n–2

(–y)

2

+

n

C

3

x

n–3

(–y)

3

+ ... +

n

C

n

(–y)

n

=

n

C

0

x

n

–

n

C

1

x

n – 1

y +

n

C

2

x

n – 2

y

2

–

n

C

3

x

n – 3

y

3

+ ... + (–1)

n

n

C

n

y

n

Thus (x–y)

n

=

n

C

0

x

n

–

n

C

1

x

n – 1

y +

n

C

2

x

n – 2

y

2

+ ... + (–1)

n

n

C

n

y

n

Using this, we have (x–2y)

5

=

5

C

0

x

5

–

5

C

1

x

4

(2y) +

5

C

2

x

3

(2y)

2

–

5

C

3

x

2

(2y)

3

+

5

C

4

x(2y)

4

–

5

C

5

(2y)

5

= x

5

–10x

4

y + 40x

3

y

2

– 80x

2

y

3

+ 80xy

4

– 32y

5

.

(ii) Taking a = 1, b = x, we obtain

(1 + x)

n

=

n

C

0

(1)

n

+

n

C

1

(1)

n – 1

x +

n

C

2

(1)

n – 2

x

2

+ ... +

n

C

n

x

n

=

n

C

0

+

n

C

1

x +

n

C

2

x

2

+

n

C

3

x

3

+ ... +

n

C

n

x

n

Thus (1 + x)

n

=

n

C

0

+

n

C

1

x +

n

C

2

x

2

+

n

C

3

x

3

+ ... +

n

C

n

x

n

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