1.2² + 2.3² + 3.4² + ...

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Solution

Let $T_{n}$ be the nth term of the given series.

Thus, we have:

$T_{n} = n {(n + 1)}^{2}$

Now, let $S_{n}$ be the sum of n terms of the given series.

Thus, we have:
$S_{n} = \sum_{k = 1}^{n} T_{k}$

$S_{n} = {\sum k}_{k = 1}^{n} {(k + 1)}^{2} = \sum_{k = 1}^{n} k (k^{2} + 1 + 2 k) = \sum_{k = 1}^{n} (k^{3} + k + 2 k^{2}) = \sum_{k = 1}^{n} k^{3} + 2 \sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} k = \frac{n^{2} {(n + 1)}^{2}}{4} + \frac{2 n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2} = \frac{n^{2} {(n + 1)}^{2}}{4} + \frac{n (n + 1) (2 n + 1)}{3} + \frac{n (n + 1)}{2} = \frac{n (n + 1)}{2} (\frac{n (n + 1)}{2} + \frac{2 (2 n + 1)}{3} + 1) = \frac{n (n + 1)}{2} (\frac{n^{2} + n}{2} + \frac{4 n + 2}{3} + 1) = \frac{n (n + 1)}{2} (\frac{3 n^{2} + 3 n + 8 n + 4 + 6}{6}) = \frac{n (n + 1)}{12} (3 n^{2} + 11 n + 10) = \frac{n (n + 1)}{12} (3 n^{2} + 6 n + 5 n + 10) = \frac{n (n + 1) (n + 2) (3 n + 5)}{12}$

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