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

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

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$\frac{n(n + 1)(3n^2 + 7n + 2)}{12}$

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(B) The $n$-th term of the series is $T_n = n^2(n + 1) = n^3 + n^2$. The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n k^3 + \sum_{k=1}^n k^2$. Using the standard summation formulas,$S_n = \left[ \frac{n(n + 1)}{2} \right]^2 + \frac{n(n + 1)(2n + 1)}{6}$. Factoring out $\frac{n(n + 1)}{2}$,we get $S_n = \frac{n(n + 1)}{2} \left[ \frac{n(n + 1)}{2} + \frac{2n + 1}{3} \right]$. Simplifying the expression inside the brackets: $\frac{3n^2 + 3n + 4n + 2}{6} = \frac{3n^2 + 7n + 2}{6}$. Thus,$S_n = \frac{n(n + 1)(3n^2 + 7n + 2)}{12}$.

The sum of the series $1^2 \cdot 2 + 2^2 \cdot 3 + 3^2 \cdot 4 + \dots$ to $n$ terms is

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