Question

The sum to $10$ terms of the series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\ldots $ is

• A. $\frac{56}{111}$
• B. $\frac{58}{111}$
• C. $\frac{55}{111}$
• D. $\frac{59}{111}$

Answer 1

The Correct Option is C

$T_r=\frac{\left(r^2+r+1\right)-\left(r^2-r+1\right)}{2\left(r^4+r^2+1\right)}$
$\Rightarrow T_r=\frac{1}{2}\left[\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right]$
$T_1=\frac{1}{2}\left[\frac{1}{1}-\frac{1}{3}\right]$
$T_2=\frac{1}{2}\left[\frac{1}{3}-\frac{1}{7}\right]$
$T_3=\frac{1}{2}\left[\frac{1}{7}-\frac{1}{13}\right]$
$⋮$
$T_{10}=\frac{1}{2}\left[\frac{1}{91}-\frac{1}{111}\right]$
$\Rightarrow \sum _{r=1}^{10}{T}_r=\frac{1}{2}\left[1-\frac{1}{111}\right]=\frac{55}{111}$

Answer 2

Using partial fractions: \[ T_r = \frac{x^2 + r + 1 - (x^2 - r + 1)}{2(r^2 + r + 1)} \] \[ T_1 = \frac{1}{2} \left[ \frac{1}{1} - \frac{1}{3} \right] \] \[ T_2 = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{7} \right] \] \[ T_{10} = \frac{1}{2} \left[ \frac{1}{29} - \frac{1}{111} \right] \] Summing over 10 terms: \[ \sum_{r=1}^{10} T_r = \frac{55}{111} \]