The sum of the series 1 cdot 3 cdot 5 + 2 cdo | vedclass

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Question

(A) The given series is $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots + n 	ext{ terms}$.The $r$-th term $T_r$ is given by the p

Vedclass · Accepted Answer

$\frac{n(n + 1)(9n^2 + 23n + 13)}{6}$

Vedclass · Answer

(A) The given series is $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots + n 	ext{ terms}$.The $r$-th term $T_r$ is given by the product of the $r$-th terms of three arithmetic progressions: $(1, 2, 3, \dots)$,$(3, 5, 7, \dots)$,and $(5, 8, 11, \dots)$.$T_r = r \cdot (2r + 1) \cdot (3r + 2) = r(6r^2 + 4r + 3r + 2) = 6r^3 + 7r^2 + 2r$.The sum of $n$ terms is $S_n = \sum_{r=1}^{n} T_r = 6 \sum r^3 + 7 \sum r^2 + 2 \sum r$.Using the standard summation formulas:$S_n = 6 \left[ \frac{n(n+1)}{2} ight]^2 + 7 \left[ \frac{n(n+1)(2n+1)}{6} ight] + 2 \left[ \frac{n(n+1)}{2} ight]$.$S_n = \frac{6n^2(n+1)^2}{4} + \frac{7n(n+1)(2n+1)}{6} + n(n+1)$.$S_n = \frac{n(n+1)}{6} [9n(n+1) + 7(2n+1) + 6]$.$S_n = \frac{n(n+1)}{6} [9n^2 + 9n + 14n + 7 + 6] = \frac{n(n+1)(9n^2 + 23n + 13)}{6}$.

The sum of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$ up to $n$ terms is:

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