Question

The sum of the series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+...$ to $n$ terms is

• A. $\frac{n(n^2+1)}{n^2+n+1}$
• B. $\frac{n(n+1)}{2(n^2+n+1)}$
• C. $\frac{n(n^2-1)}{2(n^2+n+1)}$
• D. None of these

Answer 1

Answer: (B)

$\frac{n(n+1)}{2(n^2+n+1)}$

Let $T_n$ be the $n^{th}$ term of the series

$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+...$

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

$=\frac{n}{(n^2+n+1)(n^2-n+1)}=\frac{1}{2}[\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}]$

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

Now, $\sum _{r=1}^n{T}_r=\frac{1}{2}[\frac{1}{1}-\frac{1}{1+1.2}]+\frac{1}{2}[\frac{1}{1+1.2}-\frac{1}{1+2.3}]$

$+\frac{1}{2}[\frac{1}{1+2.3}-\frac{1}{1+3.4}]+...+\frac{1}{2}[\frac{1}{1+(n-1)n}-\frac{1}{1+n(n+1)}]$

$=\frac{1}{2}[1-\frac{1}{1+n(n+1)}]=\frac{n(n+1)}{2(n^2+n+1)}$