Question

The sum of infinity of the series $\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+\frac{9}{400}+\dots \dots \infty$ is

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

Answer 1

Answer: (A)

$S_1=3,5,7,9,\dots \dots$
$T_m=2m+1$
$S_2=4,36,144,400\dots \dots$
$T_p=(p\times (p+1){)}^2$
$T_n=\frac{2n+1+n^2-n^2}{n^2(n+1{)}^2}=\frac{(n+1{)}^2-n^2}{(n+1{)}^2{n}^2}=\frac{1}{n^2}-\frac{1}{(n+1{)}^2}$
$T_1=\frac{1}{1^2}-\frac{1}{2^2}$
$T_2=\frac{1}{2^2}-\frac{1}{3^2}$
$\Lambda =T_1+T_2\dots \dots \infty =1.$