If n > 2 and α , β , γ ∈ R , then the value of S = α C 0 - α + β C 1 + α + 2 β + 2 2 γ C 2 - α + 3 β + 3 2 γ C 3 + . . . . upto n + 1 terms is equal to (where, C r denotes n C r )

Step-by-step solution

S = α C 0 - C 1 + C 2 - C 3 . . . . . . . . + β - C 1 + 2 C 2 - 3 C 3 + . . . . . . . + γ 2 2 C 2 - 3 2 C 3 + . . . . . . = α ∑ k = 0 n - 1 k C k + β ∑ k = 1 n - 1 k k C k + γ ∑ k = 2 n - 1 k k 2 C k But ,   ∑ k = 0 n - 1 k C k = 0 , ∑ k = 1 n - 1 k k C k = d d x 1 - x n x = 1 = 0 and ∑ k = 2 n - 1 k k 2 C k = ∑ k = 2 n - 1 k k k - 1 + k C k = ∑ k = 2 n - 1 k k k - 1 C k + ∑ k = 1 n - 1 k k C k + C 1 = 0 + 0 + n = n ⇒ S =