The value of n + 2 · C 0 n ⋅ 2 n + 1 - n + 1 · C 1 n ⋅ 2 n + n · C 2 n ⋅ 2 n - 1 - . . . . to n + 1 terms is equal to

Question

The value of n + 2 · C 0 n ⋅ 2 n + 1 - n + 1 · C 1 n ⋅ 2 n + n · C 2 n ⋅ 2 n - 1 - . . . . to n + 1 terms is equal to, 4, 4 n, 4 n + 1, 2 n + 2

MARKS App · Accepted Answer

T r + 1 = - 1 r n - r + 2 C r n 2 n - r + 1 = n + 2 2 n + 1 - 1 r C n r 1 2 r - 2 n + 1 - 1 r r C n r 1 2 r = n + 2 2 n + 1 - 1 r C r n 1 2 r + 2 n ⋅ n - 1 r - 1 C r - 1 n - 1 1 2 r - 1 So, sum = n + 2 2 n + 1 ∑ - 1 r C r n 1 2 r + 2 n n ∑ - 1 r - 1 C r - 1 n - 1 1 2 r - 1 = n + 2 2 n + 1 1 - 1 2 n + 2 n n 1 - 1 2 n - 1 = n + 2 2 n + 1 1 2 n + 2 n n 1 2 n - 1 = n + 2 2 + 2 n = 4 n + 4 = 4 n + 1

Search any question & find its solution

Looking for more such questions to practice?