Question

Find the sum of the infinite series $\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+\frac{1}{63}+....$

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{5}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{1}{3}$

$\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+....$

$f\left(n\right)=a{n}^2+bn+c$

$9=a+b+c+$ $18=4a+2b+c$ $30=9a+3b+c$

Solve for $a,b,c$

$a=\frac{3}{2}b=\frac{9}{2}c=3$

$\left(n\right)=\frac{3n^2}{2}+\frac{9n}{2}+3=\frac{3}{2}(n^2+3n+6$

$Curvedterm=\frac{2}{3}\frac{1}{(n^2+3n+6)}=\frac{2}{3}[\frac{1}{n+1}-\frac{1}{n+2}]$

${\sum }_{n=1}\frac{2}{3}[\frac{1}{n+1}-\frac{1}{n+2}]$

in $\sum \frac{1}{n+1}$ there is are term $\frac{1}{n+1}(n=1)$

extra in $\sum \frac{1}{n+1}$

$\frac{2}{3}\times \frac{1}{2}=\frac{1}{3}$

$A$ is correct.