If the coefficients of 2nd, 3rd and the 4th terms in the expansion of (1 + x)n are in A.P., then value of n is ______.
Options
2
7
11
14
Solution
If the coefficients of 2nd, 3rd and the 4th terms in the expansion of (1 + x)n are in A.P., then value of n is 7.
Explanation:
Given expression is (1 + x)n
(1 + x)n = nC0 + nC1x + nC2x2 + nC3x3 + … nCnxn
Here, coefficient of 2nd term = nC1
Coefficient of 3rd term = nC2
And coefficient of 4th term = nC3
Given that nC1, nC2 and nC3 are in A.P.
∴ 2 . nC2 = nC1 + nC3
⇒ `2 * (n(n - 1))/2 = n + (n(n - 1)(n - 2))/(3*2*1)`
⇒ `n(n - 1) = n + (n(n - 1)(n - 2))/6`
⇒ n – 1 = `1 + ((n - 1)(n - 2))/6`
⇒ 6n – 6 = 6 + n2 – 3n + 2
⇒ n2 – 3n – 6n + 14 = 0
⇒ n2 – 9n + 14 = 0
⇒ n2 – 7n – 2n + 14 = 0
⇒ n(n – 7) – 2(n – 7) = 0
⇒ (n – 2)(n – 7) = 0
⇒ n = 2, 7
⇒ n = 7
Whereas n = 2 is not possible