Question

If $\frac{1^4}{1.3}+\frac{2^4}{3.5}+\frac{3^4}{5.7}+...+\frac{n^4}{(2n-1).(2n+1)}=\frac{1}{48}f\left(n\right)+\frac{n}{16(2n+1)}$, then $f\left(n\right)$ is equal to

• A. $n(4n^2+3n+2)$
• B. $n(4n^2+6n+5)$
• C. $n(4n^2+5n+6)$
• D. None of these

Answer 1

The correct option is B $n(4n^2+6n+5)$

We have, $t_n=\frac{n^4}{(2n-1)(2n+1)}=\frac{n^2}{4}+\frac{1}{16}+\frac{1}{16(4n^2-1)}$

$=\frac{4n^2+1}{16}+\frac{1}{32}[\frac{1}{2n-1}-\frac{1}{2n+1}]$

$\therefore S_n=\sum _{n=1}^n{t}_n=\sum _{n=1}^n\frac{4n^2+1}{16}+\frac{1}{32}\sum _{n=1}^n[\frac{1}{2n-1}-\frac{1}{2n+1}]$

$=\frac{1}{4}\frac{n(n+1)(2n+1)}{6}+\frac{1}{16}n+\frac{1}{32}[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}]$

$=\frac{n}{48}(4n^2+6n+5)+\frac{1}{32}(1-\frac{1}{2n+1})$

$=\frac{n(4n^2+6n+5)}{48}+\frac{m}{16(2n+1)}$

$\therefore f\left(n\right)=n(4n^2+6n+5)$