Question

Let $S_k,k=1,2,3$.... denote the sum of infinite geometric series whose first term is $(k^2-1)$ and the common ratio is $\frac{1}{k}$, Find the value of $\sum _{k-1}^{\infty }\frac{S_k}{2^{k-1}}$

• A. $8$
• B. $1$
• C. $2$
• D. $5$

Answer 1

The correct option is A $8$

$S_k=(k^2-1)+\frac{k^2-1}{k}+\frac{k^2-1}{k^2}+....\infty$
Hence
$S_k=\frac{k^2-1}{1-\frac{1}{k}}$
$=k(k+1)$
$=k^2+k$
Now
$\lambda =\frac{k^2+k}{2^{k-1}}$
${\sum }_{k=1}\frac{S_k}{2^{k-1}}$
$={\sum }_{k=1}{\lambda }_k$
Hence
${\sum }_{k=1}{\lambda }_k=\alpha =2+\frac{6}{2}+\frac{12}{2^2}+\frac{20}{2^3}+\frac{30}{2^4}..\infty$
$\frac{\alpha }{2}=\frac{2}{2}+\frac{6}{2^2}+\frac{12}{2^3}+\frac{20}{2^4}...\infty$
Hence
$\alpha -\frac{\alpha }{2}=2+\frac{4}{2}+\frac{6}{2^2}+\frac{8}{2^3}+\frac{10}{2^4}...\infty$
Hence
$\frac{\alpha }{2}=2[1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}...\infty ]$
Now
$Sum=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}...\infty$
It is an AGP with common ration of 0.5.
Hence
$\frac{Sum}{2}=\frac{1}{2}+\frac{2}{4}+\frac{3}{2^3}+\frac{4}{2^4}+...\infty$
Hence
$Sum-\frac{Sum}{2}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...\infty$
$\frac{Sum}{2}=\frac{1}{1-\frac{1}{2}}=2$
Hence
$Sum=2$.
Hence
$\frac{\alpha }{2}=2[1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}...\infty ]$
$\frac{\alpha }{2}=2\left[Sum\right]$
$\frac{\alpha }{2}=2\left(2\right)=4$
Hence
$\alpha =8$.
${\sum }_{k=1}{\lambda }_k=8$
${\sum }_{k=1}\frac{S_k}{2^{k-1}}=8$