Question

If S_n = $\sum _{r=1}^n\frac{1+2+2^2+...\mathrm{Sum}\mathrm{to}r\mathrm{terms}}{2^r}$, then S_n is equal to
(a) 2ⁿ − n − 1

(b) $1-\frac{1}{2^n}$

(c) $n-1+\frac{1}{2^n}$

(d) 2ⁿ − 1

Answer 1

(c) $n-1+\frac{1}{2^n}$

We have:
S_n = $\sum _{r=1}^n\frac{1+2+2^2+...\mathrm{sum}\mathrm{to}r\mathrm{terms}}{2^r}$
$$\begin{gathered} \Rightarrow S_n=\sum _{r=1}^n\frac{1\left(2^r-1\right)}{2^r} \\ \Rightarrow S_n=\sum _{r=1}^n\left(1-\frac{1}{2^r}\right) \\ \Rightarrow S_n=n-\sum _{r=1}^n\left(\frac{1}{2^r}\right) \\ \Rightarrow S_n=n-\left[\frac{\left({\displaystyle \frac{1}{2}}\right)\left\{1-{\left({\displaystyle \frac{1}{2}}\right)}^n\right\}}{1-{\displaystyle \frac{1}{2}}}\right] \\ \Rightarrow S_n=n-\left[1-{\left(\frac{1}{2}\right)}^n\right] \\ \Rightarrow S_n=n-1+\frac{1}{2^n} \end{gathered}$$