The sum of the series
$(1 + 2) + (1 + 2 + 2^{2}) + (1 + 2 + 2^{2} + 2^{3})$

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Solution

$n^{t h}$ term $= \frac{a (r^{r + 1} - 1)}{r - 1} = \frac{1 (2^{n + 1} - 1)}{2 - 1} {.2}^{n + 1} - 1$
sum of series $= \sum_{k = 1}^{n} 2^{k + 1} - 1$
$= 2 \frac{(2^{n + 1} - 1)}{1} - 2 - n$ $[∵ \sum_{k = 0}^{n} 2^{k + 1} = 2 (2^{n + 1} - 1)]$
$= 2^{n + 2} - 4 - n$ $[∵ \sum_{k = 1}^{n} 2^{k + 1} = 2 (2^{n + 1} - 1 - 2)]$
$= 2^{n + 2} - (4 + n)$
For, $n = 3,$
$\Rightarrow$ Sum $= 2^{5} - (4 + 3)$
$= 32 - 7$
$= 25$
Hence, solved.

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The sum of the series(1+2)+(1+2+22)+(1+2+22+23)

nth term =a(rr+1−1)r−1=1(2n+1−1)2−1.2n+1−1sum of series =∑nk=12k+1−1=2(2n+1−1)1−2−n [∵∑nk=02k+1=2(2n+1−1)]=2n+2−4−n [∵∑nk=12k+1=2(2n+1−1−2)]=2n+2−(4+n)For, n=3,⇒ Sum =25−(4+3) =32−7 =25Hence, solved.

The sum of the series
$(1 + 2) + (1 + 2 + 2^{2}) + (1 + 2 + 2^{2} + 2^{3})$