The $n th$ term of a series is given by $t_n=\frac{n^5+n^3}{n^4+n^2+1}$ and if sum of its $n$ terms can be expressed as $S_n=a_n^2+a+\frac{1}{b_n^2+b}$, where $a_n$ and $b_n$ are the $n th$ terms of some arithmetic progressions and $a, b$ are some constants, prove that $\frac{b_n}{a_n}$ is a constant.

Since, $t_n=\frac{n^5+n^3}{n^4+n^2+1}$

$$\begin{gathered} =n-\frac{n}{n^4+n^2+1} \\ =n+\frac{1}{2\left(n^2+n+1\right)}-\frac{1}{2\left(n^2-n+1\right)} \end{gathered}$$

Sum of $n$ terms $S_n=\sum t_n$

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

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

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

$={\left(\frac{n}{\sqrt{2}}+\frac{1}{2\sqrt{2}}\right)}^2-\frac{1}{8}-\frac{1}{2}+\frac{1}{{\left(n\sqrt{2}+\frac{1}{\sqrt{2}}\right)}^2+\frac{3}{2}}$ 

$={\left(\frac{n}{\sqrt{2}}+\frac{1}{2\sqrt{2}}\right)}^2-\frac{5}{8}+\frac{1}{{\left(n\sqrt{2}+\frac{1}{\sqrt{2}}\right)}^2+\frac{3}{2}}$ 

but given, $S_n=a_n^2+a+\frac{1}{b_n^2+b}$

On comparing, we get $a_n=\frac{n}{\sqrt{2}}+\frac{1}{2\sqrt{2}}, a=-\frac{5}{8}, b_n=\left(n\sqrt{2}+\frac{1}{\sqrt{2}}\right), b=\frac{3}{2}$ 

$\therefore \frac{b_n}{a_n}=2$, which is constant.