Question

Find the middle terms in the expansion of:

(i) (x – 1/x)¹⁰

(ii) (1 – 2x + x²)ⁿ

(iii) (1 + 3x + 3x² + x³)²ⁿ

(iv) (2x – x²/4)⁹

(v) (x – 1/x)²ⁿ⁺¹

Answer 1

(i) (x – 1/x)¹⁰

Now, we have,

(x – 1/x)¹⁰ where, n = 10 (even number)

Therefore the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6^th term

Then,

T₆ = T₅₊₁

Thus, the middle term is -252.

(ii) (1 – 2x + x²)ⁿ

We have,

(1 – 2x + x²)ⁿ = (1 – x)²ⁿ where, n is an even number.

Therefore the middle term is (2n/2 + 1) = (n + 1)^th term.

Now,

T_n = T_n+1

= ²ⁿC_n (-1)ⁿ (x)ⁿ

= (2n)!/(n!)² (-1)ⁿ xⁿ

Thus, the middle term is (2n)!/(n!)² (-1)ⁿ xⁿ.

(iii) (1 + 3x + 3x² + x³)²ⁿ

We have,

(1 + 3x + 3x² + x³)²ⁿ = (1 + x)⁶ⁿ where, n is an even number.

Therefore the middle term is (n/2 + 1) = (6n/2 + 1) = (3n + 1)^th term.

Now,

T_2n = T_3n+1

= ⁶ⁿC_3n x³ⁿ

= (6n)!/(3n!)² x³ⁿ

Thus, the middle term is (6n)!/(3n!)² x³ⁿ.

(iv) (2x – x²/4)⁹

We have,

(2x – x²/4)⁹ where, n = 9 (odd number)

Therefore the middle terms are ((n + 1)/2) = ((9 + 1)/2) = 10/2 = 5 and

((n + 1)/2 + 1) = ((9 + 1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6

The terms are 5^th and 6^th.

Then,

T₅ = T₄₊₁

Thus, the middle term is 63/4 x¹³ and -63/32 x¹⁴.

(v) (x – 1/x)²ⁿ⁺¹

Now we have,

(x – 1/x)²ⁿ⁺¹ where, n = (2n + 1) is an (odd number)

Therefore the middle terms are ((n + 1)/2) = ((2n + 1 + 1)/2) = (2n + 2)/2 = (n + 1) and

((n + 1)/2 + 1) = ((2n + 1 + 1)/2 + 1) = ((2n + 2)/2 + 1) = (n + 1 + 1) = (n + 2)

The terms are (n + 1)^th and (n + 2)^th.

Then,

Thus, the middle term is (-1)ⁿ.²ⁿ⁺¹C_n x and (-1)ⁿ⁺¹.²ⁿ⁺¹C_n (1/x).