<s>[INST] Let $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}.... =A$. What is $\frac {1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}....$ in terms of A? Find the value of $\frac {1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}....\infty$ in terms of A
For the first expression, the $T_n$ term will be 
$$T_n=\frac{1}{\frac{((n)(n+1))^2}{4}}$$
Now $$T_n=4\left[\sum \frac{1}{(n)^2(n+1)^2}\right]$$
How should I proceed?
 [/INST] Given expression$$\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}…$$
Which is equal to 
$$\frac1{2^2}\left[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}…\right]$$
Hence the given Series sums up to $$\frac{A}4$$
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