Question

$S_1=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\cdots \infty$
$S_2=\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+\cdots \infty \text{If}{S}_1-2S_2=\frac{a}{12},\text{then}a=$

• A. $0.5$
• B. $1$
• C. $2$
• D. $2.5$

Answer 1

The correct option is C $2$

$S_1=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\cdots \infty =\sum _{n=1}^{\infty }\frac{1}{n(n+1)}=\sum _{n=1}^{\infty }(\frac{1}{n}-\frac{1}{n+1})=\underset{n\to \infty }{lim}(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{n}-\frac{1}{n+1})=\underset{n\to \infty }{lim}(1-\frac{1}{n+1})\therefore S_1=1-\underset{n\to \infty }{lim}\frac{1}{n+1}=1$

$S_2=\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+\cdots \infty =(\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+\cdots \infty )+(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+\cdots \infty )=\frac{\frac{1}{3}}{1-\frac{1}{3^2}}+\frac{\frac{1}{5^2}}{1-\frac{1}{5^2}}=\frac{3}{8}+\frac{1}{24}=\frac{5}{12}\therefore S_1-2S_2=1-2\times \frac{5}{12}=\frac{2}{12}\therefore a=2$