Question

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

$S_{n}=1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+.....+\frac{\left(2 n - 1\right)}{2^{n - 1}}$
$\frac{1}{2}S_{n}=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+.....+\frac{\left(2 n - 3\right)}{2^{n - 1}}+\frac{\left(2 n - 1\right)}{2^{n}}$
Subtracting, we get,
$\frac{1}{2}S_{n}=1+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+.....+\frac{2}{2^{n - 1}}-\frac{\left(2 n - 1\right)}{2^{n}}$
$\frac{1}{2}S_{n}=1+2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . . . \left(n - 1\right) t e r m s\right)-\frac{\left(2 n - 1\right)}{2^{n}}$
$\frac{1}{2}S_{n}=1+2\frac{\frac{1}{2} \left(1 - \left(\frac{1}{2}\right)^{n - 1}\right)}{\left(1 - \frac{1}{2}\right)}-\frac{\left(2 n - 1\right)}{2^{n}}$
$\frac{1}{2}S_{n}=1+2\left(1 - \frac{1}{2^{n - 1}}\right)-\frac{\left(2 n - 1\right)}{2^{n}}$
$=1+2-\frac{1}{2^{n - 2}}-\frac{\left(2 n - 1\right)}{2^{n}}$
$f\left(n\right)=S_{n}=6-\frac{1}{2^{n - 3}}-\frac{\left(2 n - 1\right)}{2^{n - 1}}\Rightarrow f\left(8\right)=6-\frac{1}{2^{5}}-\frac{15}{2^{7}}$
$f\left(8\right)=6-\frac{\left(4 + 15\right)}{2^{7}}=6-\frac{19}{2^{7}}$