Question

If \( s=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots \ldots \), then \( 1-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+\ldots . . . \) equals 

(A) \( \frac{s}{2} \) 

(B) \( \frac{1}{4}-s \) 

(C) \( \frac{s}{2}-2 \) 

(D) \( \frac{s-1}{2} \)

Answer 1

Correct option is (A) \(\frac s2\)

\(s = \frac 1{1^2} + \frac 1{2^2} + \frac1{3^2} + \frac 1{4^2}+....\)

Now, \(1 - \frac1{2^2} + \frac 1{3^2} - \frac1{4^2} + ....\) 

\(= \left(1 + \frac1{2^2} + \frac1{3^2} + \frac1{4^2}+ ....\right) - 2\left(\frac1{2^2} + \frac1{4^2} + \frac1{6^2}+....\right)\)

\(= s - \frac2{2^2}\left(1 + \frac1{2^2} + \frac1{3^2}+ ....\right)\)

\(=s - \frac 12s \)

\(= \frac s2\)