Find the sum of n terms of the series 1 + 6 + | vedclass

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(B) Let the series be $S_n = 1 + 6 + 13 + 22 + 33 + \dots + T_n$. The differences between consecutive terms are $5, 7, 9, 11, \dots$,which form an Ari

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$\frac{n}{6}[2n^2 + 9n - 5]$

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(B) Let the series be $S_n = 1 + 6 + 13 + 22 + 33 + \dots + T_n$. The differences between consecutive terms are $5, 7, 9, 11, \dots$,which form an Arithmetic Progression $(AP)$. Let $T_n = an^2 + bn + c$. For $n=1, T_1 = a + b + c = 1$. For $n=2, T_2 = 4a + 2b + c = 6$. For $n=3, T_3 = 9a + 3b + c = 13$. Subtracting the equations: $(4a + 2b + c) - (a + b + c) = 6 - 1 \implies 3a + b = 5$. $(9a + 3b + c) - (4a + 2b + c) = 13 - 6 \implies 5a + b = 7$. Subtracting these results: $(5a + b) - (3a + b) = 7 - 5 \implies 2a = 2 \implies a = 1$. Substituting $a=1$ into $3a + b = 5$,we get $3(1) + b = 5 \implies b = 2$. Substituting $a=1, b=2$ into $a + b + c = 1$,we get $1 + 2 + c = 1 \implies c = -2$. Thus,$T_n = n^2 + 2n - 2$. Now,$S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^2 + 2k - 2) = \sum k^2 + 2\sum k - \sum 2$. $S_n = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} - 2n$. $S_n = \frac{n(2n^2 + 3n + 1) + 6n(n+1) - 12n}{6} = \frac{n(2n^2 + 3n + 1 + 6n + 6 - 12)}{6} = \frac{n(2n^2 + 9n - 5)}{6}$.

Find the sum of $n$ terms of the series $1 + 6 + 13 + 22 + 33 + \dots$

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