## Binomial Theorem

We know that any algebraic expression with two variables is called a binomial.

Observe the following powers of the binomial (x + y).

(x + y)0 = 1

(x + y)1 = x + y

(x + y)1 = x + y

(x + y)2 = x2 + 2xy + y2

(x + y)3 = x3 + 3x2y + 3xy2 + y3

(x + y)4 = (x + y)3(x + y)

= (x3 + 3x2y + 3xy2 + y3) (x + y)

= x4 + 3x3y + 3x2y2 + xy3 + x3y + 3x2y2 + 3xy3+ y4

(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5

What do you notice in the above expansion of various powers of (x + y)?

1. The number of terms in the expansion is one more than the exponent.
2. In each expansion:
The exponent of the first term is same as the exponent of the binomial. The exponent of y in the first term is zero.
3. Subsequently, in each successive term, the exponent of x decreases by 1 with a simultaneous increase of 1 in the exponent of y.
4. The sum of the exponents of x and y in each term is equal to the exponent of the binomial.
5. The exponent of x in the last term is zero and that of y is equal to the exponent of the binomial.

#### Binomial theorem for a positive integral exponent

Theorem:

If n is a natural number, #### Example 1 #### Example 2 #### Example 3

Find the constant term or the term independent of x in the expansion of (3x – 5/x2 )9.

Solution:

Tr +1 = 9Cr (3x)9-r * (–5/x9)r

= 9Cr 39-r * x9-r * (–1)r * 5r/x2r

=9Cr 39-r* 5r (–1)r * x9-3r ———————— (i)

To get the constant term of the expansion we have to find r so that 9 – 3r = 0

⇒ r = 3

∴ 4th term is the one independent of x, i.e., the constant term.

#### Try these questions

1. Expand (x + 1/y)7
2. Find the middle term of the following expansion

3. (x/a + y/b)6
4. (√a – b)8
5. (x2/y – y2/x)8
6. (xy – 1/x2y2)5
7. (a/x + x/a)5
8. (x – 3/y)5
9. Find the term containing x5 in the expansion of (x – 1/x)11
10. Write the 14th term in the expansion of (3 + x)15
11. Write the 10th term in the expansion of (3 + x)12

1. Expand (x + 1/y)7 Solution:
(x + 1/y)7 = 7C0 x7 + 7C1 x6(1/y) + 7C2 x5 (1/y)2 + 7C3 x4 (1/y)3 + 7C4 x3(1/y)4
+ 7C5 x2 (1/y)2 + 7C6 x(1/y)6 + 7C7 (1/y)7

= x7 + 7x6(1/y) + 7C2 x5(1/y)2 + 7C3 x4(1/y)3 + 7C4 x4(1/y)4
+ 7C5 x2 (1/y)2 + 7C6 x(1/y)6 + (1/y))7
here

7!  =  7 *6 *5 *4 *3 *2 *1

7!                  7!           7 *6 *5 *4 *3 *2 *1
7C2 =  ————  =  ——  =  ——————————  =  21
(7 – 2)!2!        5!2!           5 *4 *3 *2 *1 *2 *1

7!               7!        7 * 6 * 5
7C3 =  ————  =  ——  =  ————  =  35
(7 – 3)!3!        4!3!        3 * 2 *1

7C4  =  7C3 = 35 (since nCr = nCn-r )

7C5 = 7C2 = 21
7C6 = 7C1 = 7
∴ (x + 1/y)7 = x7 + 7x6(1/y) + 21x5(1/y)2 + 35x4(1/y)3 + 35x3(1/y)4 + 21x2(1/y)5 + 7x(1/y)6 + (1/y)7

2. (x/a + y/b)6
Solution:
Expansion contains 7 terms in it and 4th term is the middle term.
T4= T1+3 = 6C3 (x/a)6-3 (y/b)3
= 6C3 (x/a)3 (y/b)3

3.  (√a – b)8
Solution:
Expansion contains 9 terms in it and 5th term is the middle term.
T5 = T1+4 = 8C4 (√a)8-4 (b)4
= 8C4 (√a)4 (b)4
= 8C4 a2b4

4. (x2/y – y2/x)8
Solution:
Expansion contains 9 terms in it and 5th term is the middle term.
T5 = T1+4 = 8C4 (x2/y)8–4 (–y2/x)4
= 8C4 (x2/y)4 (–y2/x)4
x8 *   y8
= 8C4  ————
x4 * y4
= 8C4  x4 y4

5. (xy – 1/x2 y2)5
Solution:
Expansion contains 6 terms in it and 3rd and 4th terms are the middle terms. 6. (a/x + x/a)5
Solution:
Expansion contains 5 terms in it and 3rd term is the middle term. 7. (x – 3/y)5
Solution:
Expansion contains 10 terms in it and 5th and 6th terms are the middle terms. 8. Find the term containing x5 in the expansion of (x – 1/x)11
Solution: 9. Write the 14th term in the expansion of (3 + x)15
Solution: 10. Write the 10th term in the expansion of (3 + x)12
Solution: 