Solve
$1.2 + {2.2}^{2} + {3.2}^{3} + . . . + n {.2}^{n} = (n - 1) 2^{n + 1} + 2.$

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Solution

We have,
$\begin{matrix} 1.2 + {2.2}^{2} + {3.2}^{3} + . . . . . . . . . + n {.2}^{2} = (n - 1) 2^{n + 1} + 2 1.2 + {2.2}^{2} + {3.2}^{3} + . . . . . . . . . + n {.2}^{2} \sum_{r = 1}^{n} r 2^{r} S = 1.2 + {2.2}^{2} + {3.2}^{3} + . . . . . . . . . + n {.2}^{2} S = {1.2}^{2} + {2.2}^{3} + . . . . . . . . . + (n - 1) 2^{n} + n {.2}^{n + 1} - ------------------------------------------------------- - - S = 1.2 + {1.2}^{2} + {1.2}^{3} + . . . . . . . . . + 2^{n} - n 2^{n + 1} - S = \frac{2 (1 - 2^{n})}{- 1} - n 2^{n + 1} - S = n 2^{n + 1} - 2 (2^{n + 1}) S = n 2^{n + 1} - 2 (2^{n} - 1) S = 2^{n + 1} (n - 1) + 2 \end{matrix}$

Hence Proved.

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