Question

Sum of terms;-
$S_n=\frac{1^4}{1.3}+\frac{2^4}{3.5}+\frac{3^4}{5.7}+...........n$ terms

Answer 1

$S_n=\frac{1^4}{1\cdot 3}+\frac{2^4}{3\cdot 5}+\frac{3^4}{5\cdot 7}+....nterms$

$S_n=\sum _{r=1}^n\frac{r^4}{(2r-1)(2r+1)}$

$S_n=\sum _{r=1}^n\frac{\frac{r^2(2r-1)(2r+1)}{4}+\frac{r^2}{4}}{(2r-1)(2r+1)}$

$S_n=\sum _{r=1}^n\frac{r^2}{4}+\frac{r^2}{4(2r-1)(2r+1)}$

$S_n=\sum _{r=1}^n\frac{r^2}{4}+\sum _{r=1}^n\frac{r^2}{4(2r-1)(2r+1)}$

$S_n=\frac{1}{4}\sum _{r=1}^n{r}^2+\frac{1}{4}\sum _{r=1}^n\frac{r^2}{(2r-1)(2r+1)}$

$S_n=\frac{1}{4}\frac{n(n+1)(2n+1)}{6}+\frac{1}{4}\sum _{r=1}^n\frac{r^2}{(2r-1)(2r+1)}$

$S_n=\frac{n(n+1)(2n+1)}{24}+\frac{1}{4}\sum _{r=1}^n\frac{\frac{(2r-1)(2r+1)}{4}+\frac{1}{4}}{(2r-1)(2r+1)}$

$S_n=\frac{n(n+1)(2n+1)}{24}+\frac{1}{4}\sum _{r=1}^n\frac{1}{4}+\frac{1}{4}\sum _{r=1}^n\frac{1}{4(2r-1)(2r+1)}$

$S_n=\frac{n(n+1)(2n+1)}{24}+\frac{n}{16}+\frac{1}{16}\sum _{r=1}^n\frac{(2r+1)-(2r-1)}{2(2r-1)(2r+1)}$

$S_n=\frac{n(n+1)(2n+1)}{24}+\frac{n}{16}+\frac{1}{32}\sum _{r=1}^n(\frac{1}{2r-1}-\frac{1}{2r+1})$

$S_n=\frac{n(n+1)(2n+1)}{24}+\frac{n}{16}+\frac{1}{32}(\frac{1}{1}-\frac{1}{2n+1})$

$S_n=\frac{n(n+1)(2n+1)}{24}+\frac{n}{16}+\frac{1}{32}-\frac{1}{32(2n+1)}$

$\therefore S_n=\frac{n(n+1)(2n+1)}{24}+\frac{n}{16}+\frac{1}{32}-\frac{1}{32(2n+1)}$