Find the sum of n terms of a series

Question

Find the sum of n terms of series whose $n$th term is $\frac{n^2(n^2-1)}{4n^2-1}$.

Answer 1

Hint

If you use partial fraction decomposition, you find something which is very interesting since $$\frac{n^2(n^2-1)}{4n^2-1}=\frac{n^2}{4}-\frac{3}{16}-\frac{3}{32 (2 n-1)}+\frac{3}{32 (2 n+1)}$$ Have a look at the third and fourth term and notice that $\cdots ???$