Question

If the coefficients of 2^nd, 3^rd and 4^th terms in the expansion of (1 + x)²ⁿ are in AP, show that 2n² – 9n + 7 = 0.

Answer 1

For (1 + x)²ⁿ 

a=1, b=x and N=2n 

We have, t_r+1 = (_r^N) a^N-rb^r

For the 2^nd term, r = 1

t₂ = t₁₊₂

=  (₁²ⁿ) (1)^2n-1 (X)¹

= (2n)x .....[(₁ⁿ)=n]

Therefore, the coefficient of 2^nd term = (2n) 

For the 3^rd term, r = 2

t₃ = t₂₊₁

=  (₂²ⁿ) (1)^2n-2 (X)²

= (n)(2n-1)X²

Therefore, the coefficient of 3rd term = (n)(2n-1) 

For the 4^th term, r=3

t₄ = t₃₊₁

=  (₃²ⁿ) (1)^2n-3 (X)³

Therefore, the coefficient of 3^rd term = \(\frac{2(n)(2n-1).(n-1)}{3}\)

As the coefficients of 2^nd, 3^rd and 4^th terms are in A.P. 

Therefore, 

2×coefficient of 3^rd term = coefficient of 2^nd term + coefficient of the 4^th term

2 x (n)(2n-1) = (2n) + \(\frac{2(n)(2n-1).(n-1)}{3}\)

Dividing throughout by (2n),

2n-1 = 1 + \(\frac{(2n-1)(n-1)}{3}\)

2n-1 = \(\frac{3+(2n-1)(n-1)}{3}\)

• 3 (2n-1) = 3 + (2n-1)(n-1) 

• 6n – 3 = 3 + (2n² - 2n – n + 1) 

• 6n – 3 = 3 + 2n² - 3n + 1 

• 3 + 2n² - 3n + 1 - 6n + 3 = 0 

• 2n² - 9n + 7 = 0

Conclusion : If the coefficients of 2^nd, 3^rd and 4^th terms of (1 + x)²ⁿ are in A.P. then 2n² - 9n + 7 = 0