JEE Advanced Mathematics Question (2019) — Solution

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 - 1 2 n - 2 + n 2 n - 1 = n 2 n - 2 n - 1 + 2 = n n + 1 2 n - 2 ∴ r C r   n = n C r - 1   n - 1 ∴   ∑ r = 0 n C r   n = 2 n c   ∑ k = 0 n C k   n k = ∑ k = 0 n k C k   n = ∑ k = 0 n k n k C k - 1   n - 1 = n ∑ k = 0 n C k - 1   n - 1 = n 2 n - 1 d   ∑ k = 0 n C k   n 3 k = C 0   n + C 1 3 + C 2   n 3 2 + … + C n   n 3 n   n = 1 + 3 n = 4 n now n n + 1 2 n n + 1 2 n − 2 n 2 n − 1 4 n = 0 n n + 1 2 2 n - 1 - n 2 n + 1 2 2 n - 3 = 0 2 2 n - 1 - n 2 2 n - 3 = 0 n = 4 Now ∑ k = 0 4 C k   4 k + 1 = 1 5 ∑ k = 0 4 C k + 1   5 = 1 5 2 5 - 1 ∴ C r   n r + 1 = C r + 1   n + 1 n + 1 = 31 5 = 6.20 ∴   ∑ r = 1 n C r = 2 r - 1   n