The sum to 10 terms of the series 11+1^2+1^4+21+2^2+2^4+31+3^2+3^4+ is:

Question

The sum to 10 terms of the series $\dfrac{1}{1+1^2+1^4}+\dfrac{2}{1+2^2+2^4}+\dfrac{3}{1+3^2+3^4}+\cdots$ is:

Accepted Answer

$$
\begin{aligned}
& T_n=\frac{n}{1+n^2+n^4} \\
& =\frac{n}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\right)} \\
& =\frac{1}{2}\left[\frac{\left(\mathrm{n}^2+\mathrm{n}+1\right)-\left(\mathrm{n}^2-\mathrm{n}+1\right)}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\right)}\right] \\
& \Rightarrow \mathrm{T}_{\mathrm{n}}=\frac{1}{2}\left[\frac{1}{\left(\mathrm{n}^2-\mathrm{n}+1\right)}-\frac{1}{\left(\mathrm{n}^2+\mathrm{n}+1\right)}\right] \\
& \mathrm{S}_{\mathrm{n}}=\sum_{n=1}^{10} T_n \\
& =\frac{1}{2} \sum\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right) \\
& =\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)\right. \\
& \left.\dots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+\left(\frac{1}{91}-\frac{1}{111}\right)\right]
\end{aligned}
$$

$$ \therefore $$ $ S=\frac{1}{2}\left(1-\frac{1}{111}\right)=\frac{55}{111} $

Solutions · 1