If $S_{n}=\sum_{r=1}^{n} \frac{1+2+2^{2}+\ldots \text { Sum to } r \text { terms }}{2^{r}}$, then $S_{n}$ is equal to
(a) $2^{n}-n-1$
(b) $1-\frac{1}{2^{n}}$
(c) $n-1+\frac{1}{2^{n}}$
(d) $2^{n}-1$
(c) $n-1+\frac{1}{2^{n}}$
We have:
$S_{n}=\sum_{r=1}^{n} \frac{1+2+2^{2}+\ldots \text { sum to } r \text { terms }}{2^{r}}$
$\Rightarrow S_{n}=\sum_{r=1}^{n} \frac{1\left(2^{r}-1\right)}{2^{r}}$
$\Rightarrow S_{n}=\sum_{r=1}^{n}\left(1-\frac{1}{2^{r}}\right)$
$\Rightarrow S_{n}=n-\sum_{r=1}^{n}\left(\frac{1}{2^{r}}\right)$
$\Rightarrow S_{n}=n-\left[\frac{\left(\frac{1}{2}\right)\left\{1-\left(\frac{1}{2}\right)^{n}\right\}}{1-\frac{1}{2}}\right]$
$\Rightarrow S_{n}=n-\left[1-\left(\frac{1}{2}\right)^{n}\right]$
$\Rightarrow S_{n}=n-1+\frac{1}{2^{n}}$
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