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

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(B) The given series is $S = 3 \cdot 6 + 4 \cdot 7 + 5 \cdot 8 + \dots$ up to $(n - 2)$ terms.Let the $k$-th term of the series be $T_k = (k + 2)(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 $k$-th term of the series be $T_k = (k + 2)(k + 5) = k^2 + 7k + 10$.The sum of $m$ terms is $\sum_{k=1}^{m} (k^2 + 7k + 10)$,where $m = n - 2$.Sum $= \sum_{k=1}^{m} k^2 + 7 \sum_{k=1}^{m} k + \sum_{k=1}^{m} 10 = \frac{m(m+1)(2m+1)}{6} + \frac{7m(m+1)}{2} + 10m$.Substituting $m = n - 2$:Sum $= \frac{(n-2)(n-1)(2n-3)}{6} + \frac{7(n-2)(n-1)}{2} + 10(n-2)$.$= \frac{(n-2)}{6} [(n-1)(2n-3) + 21(n-1) + 60] = \frac{(n-2)}{6} [2n^2 - 5n + 3 + 21n - 21 + 60] = \frac{(n-2)(2n^2 + 16n + 42)}{6} = \frac{2n^3 + 12n^2 + 10n - 84}{6}$.

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

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