The sum to 10 terms of the series 1 1 + 1 2 + 1 4 + 2 1 + 2 2 + 2 4 + 3 1 + 3 2 + 3 4 + . . . is [2023]

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Solution

Q.
The sum to 10 terms of the series $\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + . . .$ is [2023]

1 $\frac{56}{111}$

2 $\frac{58}{111}$

3 $\frac{55}{111}$

4 $\frac{59}{111}$

Ans.
(3)

$\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + \dots + up to 10 terms$

$\sum_{r = 1}^{10} \frac{r}{1 + r^{2} + r^{4}} = \frac{1}{2} \sum_{r = 1}^{10} \frac{2 r}{1 + r^{2} + r^{4}}$

$= \frac{1}{2} \sum_{r = 1}^{10} \frac{(r^{2} + r + 1) - (r^{2} - r + 1)}{(r^{2} + r + 1) (r^{2} - r + 1)}$

$= \frac{1}{2} [\sum_{r = 1}^{10} (\frac{1}{r^{2} - r + 1} - \frac{1}{r^{2} + r + 1})]$ $= \frac{1}{2} [1 - \frac{1}{10^{2} + 10 + 1}]$

$= \frac{1}{2} [1 - \frac{1}{100 + 10 + 1}] = \frac{1}{2} [1 - \frac{1}{111}] = \frac{1}{2} \times \frac{110}{111} = \frac{55}{111}$

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