Question

The sum to $n$ terms of the series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\dots \dots$ is $\frac{1}{2}-\frac{1}{a(n^b+n^c+d{n}^4+1)}$, then $a+b+c+d=$

• A. $5$
• B. $8$
• C. $4$
• D. $6$

Answer 1

The correct option is A $5$

Let $T_r$ be the $r^{th}$ term of the given series. Then,
$T_r=\frac{r}{1+r^2+r^4}=\frac{r}{(r^2+1{)}^2-r^2}$
$=\frac{r}{(r^2-r+1)(r^2+r+1)}$
$=\frac{1}{2}[\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}]$
Therefore sum of the series is
$\sum _{r=1}^n{T}_r=\frac{1}{2}\left[\sum _{r=1}^n(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1})\right]$
$=\frac{1}{2}(1-\frac{1}{3})+\frac{1}{2}(\frac{1}{3}-\frac{1}{7})+\frac{1}{2}(\frac{1}{7}-\frac{1}{13})+\frac{1}{2}(⋮⋮)+\frac{1}{2}(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1})$
$\therefore S_n=\frac{1}{2}[1-\frac{1}{n^2+n+1}]$
Hence $a=2,b=2,c=1,d=0$