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Find the sum of series up to 6^th term whose n^th term is given by n^2 + 3^n.(a) 91(b) 1284(c) 1183(d) 1092I had been asked this question in exam.I'd like to ask this question from Sum to n Terms of Special Series topic in division Sequences and Series of Mathematics – Class 11

Answer»

Right choice is (C) 1183

To explain: Given, n^th TERM is n^2 + 3^n

So, AK = k^2 + 3^k

Taking SUMMATION from k=1 to k=n on both sides, we get

\(\sum_{i=0}^na_k = \sum_{i=0}^nk^2 + \sum_{i=0}^n3^k\)

\(\sum_{i=0}^nk^2 = n(n+1) (2n+1)/6\)

\(\sum_{i=0}^n3^k = 3*(3^n-1)/ (3-1) = (3/2) (3^n-1)\)

So, \(\sum_{i=0}^na_k = n(n+1) (2n+1)/6 + (3/2) (3^n-1)\)

SUM up to 6^th term = 6*7*13/6 + (3/2) (3^6-1) = 91+1092 = 1183.


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