Find the sum of the series 2 + 4 + 7 + 11 + 1 | vedclass

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(B) Let the series be $S_n = 2 + 4 + 7 + 11 + 16 + \dots + T_n$. The differences between consecutive terms are $2, 3, 4, 5, \dots$,which form an arith

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the series be $S_n = 2 + 4 + 7 + 11 + 16 + \dots + T_n$. The differences between consecutive terms are $2, 3, 4, 5, \dots$,which form an arithmetic progression. Let $T_n = an^2 + bn + c$. For $n=1, T_1 = a + b + c = 2$. For $n=2, T_2 = 4a + 2b + c = 4$. For $n=3, T_3 = 9a + 3b + c = 7$. Subtracting the equations,we get $3a + b = 2$ and $5a + b = 3$. Solving these,$2a = 1 \implies a = 1/2$,$b = 1/2$,and $c = 1$. Thus,$T_n = \frac{1}{2}n^2 + \frac{1}{2}n + 1 = \frac{n^2 + n + 2}{2}$. The sum $S_n = \sum_{k=1}^n T_k = \frac{1}{2} [\sum k^2 + \sum k + \sum 2]$. $S_n = \frac{1}{2} [\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + 2n]$. $S_n = \frac{n}{12} [(n+1)(2n+1) + 3(n+1) + 12] = \frac{n}{12} [2n^2 + 3n + 1 + 3n + 3 + 12] = \frac{n}{12} [2n^2 + 6n + 16] = \frac{n}{6} (n^2 + 3n + 8)$.

Find the sum of the series $2 + 4 + 7 + 11 + 16 + \dots$ up to $n$ terms.

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