Suppose det ⁡ ∑ k = 0 n k ∑ k = 0 n C k   n k 2 ∑ k = 0 n C k   n k ∑ k = 0 n C k   n 3 k = 0 , holds for some positive integer n . Then ∑ k = 0 n C k   n k + 1 equals

Step-by-step solution

As given ∑ k = 0 n k ∑ k = 0 n C   n k k 2 ∑ k = 0 n C   n k k ∑ k = 0 n C   n k 3 k = 0 a   ∑ k = 0 n k = n n + 1 2 b   ∑ k = 0 n C k   n k 2 = ∑ k = 0 n k 2 - k + k n C k = ∑ k = 0 n k 2 - k C k   n + ∑ k = 0 n k C k   n = ∑ k = 0 n k k - 1 n k . n - 1 k - 1 C k - 2   n - 2 + ∑ k = 0 n k n k C k - 1   n - 1 = n n - 1 ∑ k = 0 n C k - 2   n - 2 + n ∑ k = 0 n C k - 1   n - 1 = n n