If S_n= ^ n _ r=1 1+2+2^2+ ... Sum to r terms 2^ r , then S_n is equal to:

If S_n= ^ n _ r=1 1+2+2^2+ ... Sum to r terms 2^ r , then S_n is …

MCQ

If $\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1+2+2^2+\ ...\text{ Sum to r terms}}{2^{\text{r}}},$ then $S_n$ is equal to:

A
$2^{\text{n}}-\text{n}-1$
B
$1-\frac{1}{2^{\text{n}}}$
✓
$\text{n}-1-\frac{1}{2^{\text{n}}}$
D
${2^{\text{n}}}-1$

✓

Answer

Correct option: C.

$\text{n}-1-\frac{1}{2^{\text{n}}}$

$\text{n}-1-\frac{1}{2^{\text{n}}}$

Solution:
We have,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1+2+2^2+\ ...\text{ Sum to r terms}}{2^{\text{r}}}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1(2^{\text{r}}-1)}{2\text{r}}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\Big(1-\frac{1}{2^{\text{r}}}\Big)$
$\Rightarrow\text{S}_\text{n}=\text{n}-\sum\limits^{\text{n}}_{\text{r}=1}\Big(\frac{1}{2^{\text{r}}}\Big)$
$\Rightarrow\text{S}_\text{n}=\text{n}-\Bigg[\frac{\big(\frac{1}{2}\big)\big\{1-\big(\frac{1}{2}\big)^\text{n}\big\}}{1-\frac{1}{2}}\Bigg]$
$\Rightarrow\text{S}_\text{n}=\text{n}-\Big[1-\Big(\frac{1}{2}\Big)^\text{n}\Big]$
$\Rightarrow\text{S}_\text{n}=\text{n}-1+\frac{1}{2^\text{n}}$

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