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(A) The $n$-th term of the series is $T_n = n(2n + 1)(3n + 2)$.$T_n = n(6n^2 + 4n + 3n + 2) = 6n^3 + 7n^2 + 2n$.Sum $S_n = \sum_{k=1}^{n} T_k = 6\sum

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

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(A) The $n$-th term of the series is $T_n = n(2n + 1)(3n + 2)$.$T_n = n(6n^2 + 4n + 3n + 2) = 6n^3 + 7n^2 + 2n$.Sum $S_n = \sum_{k=1}^{n} T_k = 6\sum n^3 + 7\sum n^2 + 2\sum n$.Using standard summation formulas:$S_n = 6 \left[ \frac{n(n+1)}{2} \right]^2 + 7 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 2 \left[ \frac{n(n+1)}{2} \right]$.$S_n = \frac{n(n+1)}{2} \left[ 3n(n+1) + \frac{7(2n+1)}{3} + 2 \right]$.$S_n = \frac{n(n+1)}{6} [9n^2 + 9n + 14n + 7 + 6] = \frac{n(n+1)(9n^2 + 23n + 13)}{6}$.

What is the sum of $n$ terms of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$?

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