The sum of the series 3.6 + 4.7 + 5.8 + dots | vedclass

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Question

(B) The given series is $S = 3 \cdot 6 + 4 \cdot 7 + 5 \cdot 8 + \dots$ up to $(n - 2)$ terms.Let the general term be $T_k = (k + 2)(k + 5)$ for $k =

Vedclass · Accepted Answer

$\frac{1}{6}(2n^3 + 12n^2 + 10n - 84)$

Vedclass · Answer

(B) The given series is $S = 3 \cdot 6 + 4 \cdot 7 + 5 \cdot 8 + \dots$ up to $(n - 2)$ terms.Let the general term be $T_k = (k + 2)(k + 5)$ for $k = 1, 2, \dots, (n - 2)$.Alternatively,we can write the sum as $S = \sum_{k=1}^{n} k(k+3) - (1 \cdot 4 + 2 \cdot 5) = \sum_{k=1}^{n} (k^2 + 3k) - (4 + 10)$.Using the summation formulas $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:$S = \frac{n(n+1)(2n+1)}{6} + 3 \cdot \frac{n(n+1)}{2} - 14$.$S = \frac{2n^3 + 3n^2 + n + 9n^2 + 9n - 84}{6} = \frac{2n^3 + 12n^2 + 10n - 84}{6}$.Thus,the correct option is $B$.

The sum of the series $3.6 + 4.7 + 5.8 + \dots$ up to $(n - 2)$ terms is:

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