The sum of the series 1 1 - 3 . 1 2 + 1 4 + 2 1 - 3 . 2 2 + 2 4 + 3 1 - 3 . 3 2 + 3 4 + … up to 10-terms is [2024]

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Solution

Q.
The sum of the series $\frac{1}{1 - 3 . 1^{2} + 1^{4}} + \frac{2}{1 - 3 . 2^{2} + 2^{4}} + \frac{3}{1 - 3 . 3^{2} + 3^{4}} + \dots$ up to 10-terms is [2024]

1 $- \frac{45}{109}$

2 $\frac{55}{109}$

3 $- \frac{55}{109}$

4 $\frac{45}{109}$

Ans.
(3)

Given $\frac{1}{1 - 3 . 1^{2} + 1^{4}} + \frac{2}{1 - 3 . 2^{2} + 2^{4}} + \frac{3}{1 - 3 . 3^{2} + 3^{4}} + \dots$

$T_{r} = \frac{r}{1 - 3 r^{2} + r^{4}}$

$\Rightarrow T_{r} = \frac{r}{(r^{4} - 2 r^{2} + 1) - r^{2}} = \frac{r}{{(r^{2} - 1)}^{2} - r^{2}}$

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

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

$\Rightarrow T_{r} = \frac{1}{2} [\frac{1}{r^{2} - r - 1} - \frac{1}{r^{2} + r - 1}]$

Sum of 10 terms

$\sum_{r = 1}^{10} T_{r} = \frac{1}{2} {[\frac{1}{- 1} - \frac{1}{1}]} + \frac{1}{2} [\frac{1}{1} - \frac{1}{5}] + \dots + \frac{1}{2} [\frac{1}{89} - \frac{1}{109}]$

Using telescopic,

$\Rightarrow \frac{1}{2} [- 1 - \frac{1}{109}] = \frac{1}{2} [\frac{- 109 - 1}{109}] = \frac{1}{2} [\frac{- 110}{109}] = \frac{- 55}{109}$

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