Find the sum to $n -$ terms $1 + 4 + 10 + 22 + . . .$

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Solution

Let $S = 1 + 4 + 10 + 22 + . . . + T_{n}$ (i)
$S = 1 + 4 + 10 + . . . + T_{n - 1} + T_{n}$ (ii)
(i)-(ii) $\Rightarrow$ $T_{n} = 1 + (3 + 6 + 12 + . . . + T_{n} - T_{n - 1})$
$T_{n} = 1 + 3 (\frac{2^{n - 1} - 1}{2 - 1})$
$T_{n} = {3.2}^{n - 1} - 2$
So $S = \sum T_{n} = 3 \sum 2^{n - 1} - \sum 2$
$= 3. (\frac{2^{n} - 1}{2 - 1}) - 2 n = {3.2}^{n} - 2 n - 3$

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