Find the sum of the first n terms of the seri | vedclass

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(A) The given series is $3+7+13+21+31+\ldots$Let $a_n$ be the $n^{th}$ term of the series.The differences between consecutive terms are $4, 6, 8, 10,

Vedclass · Accepted Answer

$\frac{n}{3}(n^2+3n+5)$

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(A) The given series is $3+7+13+21+31+\ldots$Let $a_n$ be the $n^{th}$ term of the series.The differences between consecutive terms are $4, 6, 8, 10, \ldots$,which form an arithmetic progression.Thus,$a_n = a_1 + \sum_{k=1}^{n-1} (4 + (k-1)2) = 3 + \sum_{k=1}^{n-1} (2k+2) = 3 + 2\frac{(n-1)n}{2} + 2(n-1) = 3 + n^2 - n + 2n - 2 = n^2 + n + 1$.Now,the sum of the first $n$ terms is $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k^2 + k + 1)$.$S_n = \sum_{k=1}^n k^2 + \sum_{k=1}^n k + \sum_{k=1}^n 1$.$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n$.$S_n = \frac{n}{6} [ (n+1)(2n+1) + 3(n+1) + 6 ]$.$S_n = \frac{n}{6} [ 2n^2 + 3n + 1 + 3n + 3 + 6 ]$.$S_n = \frac{n}{6} [ 2n^2 + 6n + 10 ]$.$S_n = \frac{n}{3} [ n^2 + 3n + 5 ]$.

Find the sum of the first $n$ terms of the series: $3+7+13+21+31+\ldots$

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