Question

If the sum of the series $1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+....+\frac{\left(2n-1\right)}{{\left(2\right)}^{n-1}}$ is $f\left(n\right),$ then the value of $f\left(8\right)$ is

• A. $4+\frac{12}{2^5}$
• B. $5+\frac{13}{2^7}$
• C. $6-\frac{19}{2^7}$
• D. $5-\frac{13}{2^7}$

Answer 1

Answer: (C)

$S_n=1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+.....+\frac{\left(2n-1\right)}{2^{n-1}}$
$\frac{1}{2}{S}_n=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+.....+\frac{\left(2n-3\right)}{2^{n-1}}+\frac{\left(2n-1\right)}{2^n}$
Subtracting, we get,
$\frac{1}{2}{S}_n=1+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+.....+\frac{2}{2^{n-1}}-\frac{\left(2n-1\right)}{2^n}$
$\frac{1}{2}{S}_n=1+2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....\left(n-1\right)terms\right)-\frac{\left(2n-1\right)}{2^n}$
$\frac{1}{2}{S}_n=1+2\frac{\frac{1}{2}\left(1-{\left(\frac{1}{2}\right)}^{n-1}\right)}{\left(1-\frac{1}{2}\right)}-\frac{\left(2n-1\right)}{2^n}$
$\frac{1}{2}{S}_n=1+2\left(1-\frac{1}{2^{n-1}}\right)-\frac{\left(2n-1\right)}{2^n}$
$=1+2-\frac{1}{2^{n-2}}-\frac{\left(2n-1\right)}{2^n}$
$f\left(n\right)=S_n=6-\frac{1}{2^{n-3}}-\frac{\left(2n-1\right)}{2^{n-1}}\Rightarrow f\left(8\right)=6-\frac{1}{2^5}-\frac{15}{2^7}$
$f\left(8\right)=6-\frac{\left(4+15\right)}{2^7}=6-\frac{19}{2^7}$