Question

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

• A. $S_n=\frac{n(n+1)}{(n^2+n+1)}$
• B. $S_n=\frac{(n+1)}{(n^2+n+1)}$
• C. $S_n=\frac{(n+1)}{2(n^2+n+1)}$
• D. $S_n=\frac{n(n+1)}{2(n^2+n+1)}$

Answer 1

The correct option is D $S_n=\frac{n(n+1)}{2(n^2+n+1)}$

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

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

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

$2a_1=\frac{1}{1^2-1+1}-\frac{1}{1^2+1+1}=\frac{1}{1}-\frac{1}{3}$

$2a_2=\frac{1}{2^2-2+1}-\frac{1}{2^2+2+1}=\frac{1}{3}-\frac{1}{7}$

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$2a_n=\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}$

Adding all the above we get

$2S_n=1-\frac{1}{n^2+n+1}=\frac{n^2+n}{n^2+n+1}$

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