Question

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, then $\frac{b_n}{a_n}$ equal to

• A. $n\sqrt{2}$
• B. $\frac{n}{\sqrt{2}}$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

Answer: (D)

$2$

Since,

$t_n=\frac{n^5+n^3}{n^4+n^2+1}=n-\frac{n}{n^4+n^2+1}=n+\frac{1}{2(n^2+n+1)}-\frac{1}{2(n^2-n+1)}$

$\therefore$ sum of n terms $S_n={\sum }_{n=1}^n{t}_n$

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

$=(1+2+3+....+n)+\frac{1}{2}\{\frac{1}{3}-1+\frac{1}{7}-\frac{1}{3}+\frac{1}{13}-\frac{1}{7}+.....+\frac{1}{n^2+n+1}-\frac{1}{n^2-n+1}\}$

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

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

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

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

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