Question

If in the expansion of (1 + x)²ⁿ coefficient of 2^nd, 3^nd and 4^th terms are in A.P. then prove that 2n² – 9n + 7 = 0.

Answer 1

In the expansion of (1 + x)²ⁿ, coefficient of 2^nd, 3^rd and 4^th terms are

T₂= T₁ + 1= ²ⁿC₁

T₃ = T₂ + 1 = ²ⁿC₂

T_{4  =} T₃ +1 =  ²ⁿC₃

According to questions,

⇒ 6 (2n – 1) = 6 + 2(2n – 1) (n – 1)

⇒ 12n – 6 = 6 + 2(2n² – 3n + 1)

⇒ 4 n² – 6n + 2 – 12n + 6 + 6 = 0

⇒ 4n² – 18n +14 = 0

⇒ 2n² – 9n + 7 = 0.

Hence Proved.