The sum to n terms of the series 1 cdot 3^2 + | vedclass

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(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.Expanding this,we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.The sum of $n$ te

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$\frac{n}{6}(n + 1)(6n^2 + 14n + 7)$

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(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.Expanding this,we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (4k^3 + 4k^2 + k)$.Using the standard summation formulas $\sum k^3 = \frac{n^2(n+1)^2}{4}$,$\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum k = \frac{n(n+1)}{2}$:$S_n = 4 \left[ \frac{n^2(n+1)^2}{4} \right] + 4 \left[ \frac{n(n+1)(2n+1)}{6} \right] + \frac{n(n+1)}{2}$.$S_n = n^2(n+1)^2 + \frac{2n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$.Taking $\frac{n(n+1)}{6}$ as a common factor:$S_n = \frac{n(n+1)}{6} [6n(n+1) + 4(2n+1) + 3]$.$S_n = \frac{n(n+1)}{6} [6n^2 + 6n + 8n + 4 + 3] = \frac{n(n+1)}{6} (6n^2 + 14n + 7)$.

The sum to $n$ terms of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ is

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