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

Question

Let \[ S=\frac{1}{2!5!}+\frac{1}{3!2!3!}+\frac{1}{5!2!1!}+\cdots \text{ up to 13 terms}. \] If $13S=\dfrac{2^k}{n!}$, $k\in\mathbb{N}$, then $n+k$ is equal to

Answer 1

To find the sum of the first 10 terms of the given series, we consider the pattern of the terms in the series:

The general term of the series can be expressed as:

$\frac{n}{1 + n^2 + n^4}$

where $ n $ is a positive integer. We need to calculate the sum $ S_{10} $, which is:

$\frac{1}{1 + 1^2 + 1^4}$

+

$\frac{2}{1 + 2^2 + 2^4}$

+

$\frac{3}{1 + 3^2 + 3^4}$

+\cdots +

$\frac{10}{1 + 10^2 + 10^4}$

We simplify the denominator of each term separately:

For $ n = 1 $, the denominator is $ 1 + 1^2 + 1^4 = 1 + 1 + 1 = 3 $ so the term is $\frac{1}{3}$.
For $ n = 2 $, the denominator is $ 1 + 2^2 + 2^4 = 1 + 4 + 16 = 21 $, so the term is $\frac{2}{21}$.
For $ n = 3 $, the denominator is $ 1 + 3^2 + 3^4 = 1 + 9 + 81 = 91 $, so the term is $\frac{3}{91}$.
Similarly compute for $ n = 4, 5, \ldots, 10 $.

After computing each term, the first few terms become:

$\frac{1}{3}$
$\frac{2}{21}$
$\frac{3}{91}$

Continue calculating until the term for $ n = 10 $:

For $ n=10 $, the denominator is $ 1 + 10^2 + 10^4 = 1 + 100 + 10000 = 10101 $, so the term is $\frac{10}{10101}$.

To find the total sum, add these fractions:

The sum can be computed as:

$S_{10} = \frac{1}{3} + \frac{2}{21} + \frac{3}{91} + \ldots + \frac{10}{10101}$

After calculating, the sum $ S_{10} $ simplifies to:

$S_{10} = \frac{55}{111}$

Thus, the correct answer is $\frac{55}{111}$.

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

The Correct Option is C

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