${1.3}^{2} + {2.5}^{2} + {3.7}^{2} + . . . .$ to 20 terms.

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Solution

General term $T_{r} = r (3 + (r - 1) 2)^{2} = r (2 r + 1)^{2} = 4 r^{3} + 4 r^{2} + r$
$∴$ Sum $= n \sum r = 1 T_{r} = 4 n \sum r = 1 r^{3} + 4 n \sum r = 1 r^{2} + n \sum r = 1 r$
$= 4 (\frac{n (n + 1)}{2})^{2} + 4 \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}$
$= \frac{1}{6} n (n + 1) (6 n^{2} + 14 n + 7)$
By putting $n = 20$
we get required sum $= \frac{1}{6} 20 (20 + 1) (6 \times 20^{2} + 14 \times 20 + 7)$ $= 188090$

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