The sum to n terms of the series: {1}{1+1²+1^{4}}+{2}{1+2²+2^{4}}+{3}{1+3²+3^{4}}+ is

Correct Answer is : n2+n2(n2+n+1){n²+n}{2(n²+n+1)}2(n2+n+1)n2+n

Correct Answer is : n2+n2(n2+n+1){n²+n}{2(n²+n+1)}2(n2+n+1)n2+n

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Manipal Entrance Test 2013 Math Paper

Question : 2 of 80

Marks: +1, -0

\frac{1}{1+1^{2}+1^{4}}+\frac{2}{1+2^{2}+2^{4}}+\frac{3}{1+3^{2}+3^{4}}+\dots

Solution:

Let

T_r

be the r

^{\text{th}}

term of the given series.

Then,

T_{r}=\frac{r}{1+r^{2}+r^{4}}r=1,2,3,\dots,n

=\frac{r}{(r^{2}+r+1)(r^{2}-r+1)}

=\frac{1}{2}\left[\frac{1}{r^{2}-r+1}-\frac{1}{r^{2}+r+1}\right]

∴ Sum of the series

=\sum\limits_{r=1}^{n} T_{r}

=\frac{1}{2}\left\{\sum\limits_{r=1}^{n}\left(\frac{1}{r^{2}-r+1}-\frac{1}{r^{2}+r+1}\right)\right\}

=\frac{1}{2}\left\{\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\dots+\left(\frac{1}{n^{2}-n+1}-\frac{1}{n^{2}+n+1}\right)\right\}

=\frac{1}{2}\left\{1-\frac{1}{n^{2}+n+1}\right\} = \frac{n^{2}+n}{2(n^{2}+n+1)}

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