Question

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

Answer 1

$T_n=\frac{n^2(n^2-1)}{4n^2-1}=\frac{n^2}{4}\cdot \frac{4n^2-4}{4n^2-1}=\frac{n^2}{4}-\frac{3}{4}\cdot \frac{n^2}{4n^2-1}=\frac{n^2}{4}-\frac{3}{16}\cdot \frac{4n^2}{4n^2-1}$

$=\frac{n^2}{4}-\frac{3}{16}-\frac{3}{16(4n^2-1)}$

$=\frac{n^2}{4}-\frac{3}{16}-\frac{3}{16(2n-1)(2n+1)}$

Therefore, $S_n=\sum T_n=\frac{n(n+1)(2n+1)}{24}-\frac{3n}{16}-\frac{3n}{16(2n+1)}$

$=\frac{n(n+1)(2n+1)}{24}-\frac{3n(n+1)}{8(2n+1}$

$=\frac{(n-1)n(n+1)(n+2)}{6(2n+1)}$