Question

The sum of the first 20 terms of the series $1+\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}=....,$ is :

Answer 1

$1+\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+.....20$ terms

General term

$T_r=\frac{2^r-1}{2^{r-1}}$

$T_1+T_2+...\dots .+T_{20}=\sum _{r=1}^{20}\frac{2^r-1}{2^{r-1}}=\sum _{r=1}^{20}=\frac{2^r-1}{\frac{2^r}{2}}$

$=2\sum _{r=1}^{20}\left(\frac{2^r-1}{2^r}\right)=2\left[\sum _{r=1}^{20}(1-\frac{1}{2^r})\right]$

$=r\left[\sum _{r=1}^{20}\left(1\right)-\sum _{r=1}^{20}\frac{1}{2^r}\right]$

$=2[20-(\frac{1}{2}+\frac{1}{2^2}+...\dots +\frac{1}{2^{20}})]$

$=40-(1+\frac{1}{2}+.....+\frac{1}{2^{19}})$

$=40-1\left(\frac{1-{\left(\frac{1}{2}\right)}^{20}}{1-\frac{1}{2}}\right)$ Sum of $20$ terms of a G.P

$=40-2(1-{\left(\frac{1}{2}\right)}^{20})$

$=38+{\left(\frac{1}{2}\right)}^{19}$.