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(B) Let the $n$-th term be $T_n$. The sequence is $2, 4, 7, 11, 16, \dots$The differences between consecutive terms are $2, 3, 4, 5, \dots$,which form

Vedclass · Accepted Answer

$\frac{n}{6}(n^2 + 3n + 8)$

Vedclass · Answer

(B) Let the $n$-th term be $T_n$. The sequence is $2, 4, 7, 11, 16, \dots$The differences between consecutive terms are $2, 3, 4, 5, \dots$,which form an arithmetic progression.Thus,$T_n = T_1 + \sum_{k=1}^{n-1} (k+1) = 2 + \frac{(n-1)(2 + n)}{2} = 2 + \frac{n^2 + n - 2}{2} = \frac{4 + n^2 + n - 2}{2} = \frac{n^2 + n + 2}{2}$.Now,the sum $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k^2 + k + 2}{2} = \frac{1}{2} \left[ \sum k^2 + \sum k + \sum 2 \right]$.$S_n = \frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + 2n \right]$.$S_n = \frac{n}{12} \left[ (n+1)(2n+1) + 3(n+1) + 12 \right] = \frac{n}{12} \left[ 2n^2 + 3n + 1 + 3n + 3 + 12 \right]$.$S_n = \frac{n}{12} (2n^2 + 6n + 16) = \frac{n}{6} (n^2 + 3n + 8)$.

$2 + 4 + 7 + 11 + 16 + \dots$ to $n$ terms =

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