Question

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

• A. $1$
• B. $2$
• C. $32$
• D. $0$

Answer 1

Answer: (A)

Let $S=\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+\frac{9}{400}+\dots \infty$
$=\frac{3}{1^2\cdot 2^2}+\frac{5}{2^2\cdot 3^2}+\frac{7}{3^2\cdot 4^2}+\frac{9}{4^2\cdot 5^2}+\dots \infty$
$=\frac{2^2-1^2}{1^2\cdot 2^2}+\frac{3^2-2^2}{2^2\cdot 3^2}+\frac{4^2-3^2}{3^2\cdot 4^2}+\frac{5^2-4^2}{4^2\cdot 5^2}+\dots \infty$
$=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{5^2}+\dots \infty$
$=1$