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}+\ldots\infty $
$=\frac{3}{1^{2}\cdot2^{2}}+\frac{5}{2^{2} \cdot3^{2}}+\frac{7}{3^{2} \cdot4^{2}}+\frac{9}{4^{2} \cdot5^{2}}+\ldots\infty $
$=\frac{2^{2}-1^{2}}{1^{2}\cdot2^{2}}+\frac{3^{2}-2^{2}}{2^{2}\cdot3^{2}}+\frac{4^{2}-3^{2}}{3^{2} \cdot4^{2}}+\frac{5^{2} -4^{2}}{4^{2}\cdot5^{2}}+\ldots\infty$
$=1-\frac{1}{2^{2}}+\frac{1}{2^{2}}-\frac{1}{3^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+\frac{1}{4^{2}}-\frac{1}{5^{2}}+\ldots\infty$
$=1$