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Find the value of the series $$\frac{8}{5} + \frac{16}{65} + \frac{24}{325} + \frac{32}{1025} + ...$$ to 20 terms.

The numerator seems to follow a clearly defined pattern here: it is an AP with common difference of 8. However, I cannot seem to find a pattern for the denominator here. The ratios are $13, 5, 3.15$ etc. whereas the differences are $60, 260, 700$. Even the double differences are not seemingly following a pattern. One last thing I noticed was that $5 = 2^2+1$, $65 = 8^2+1$, $325 = 18^2+1$, $1025 = 32^2+1$. The numbers $2,6,18,32$ are again not in an obvious pattern.

What is the answer to the problem? Finding the nth term is also good enough, as I should be able to take it from there.

Note: I read this question, and the topic behind these kind of problems seems to involve the calculus of finite differences. If someone could also point to some resources regarding this topic, I would read up more on this as well.

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  • $\begingroup$ impossible without more context $\endgroup$ – FearfulSymmetry Aug 12 '20 at 8:45
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    $\begingroup$ @ParamanandSingh this was from a test paper I was attempting. I haven't made a typo copying the question, although the problem setters may have made one. I'll try to dig up a solution (if I can find one) and update the question if that adds more "context". $\endgroup$ – Aniruddha Deb Aug 12 '20 at 8:50
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    $\begingroup$ The denominator seems to be $4n^4+1$. $\endgroup$ – Jaap Scherphuis Aug 12 '20 at 8:50
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    $\begingroup$ If you're not sure, you can always use Lagrange interpolation, which graphing software (e.g Desmos) can do. $\endgroup$ – Toby Mak Aug 12 '20 at 8:52
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    $\begingroup$ Using your observation that it was one more than the squares of 2,8,18,32, I saw the same as Toby Mak that these were twice the squares 1,4,9,16. This gives $((2n)^2)^2+1=4n^2+1$. $\endgroup$ – Jaap Scherphuis Aug 12 '20 at 8:56
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I believe your sum was intended to be

$$\sum \frac{8k}{4k^4+1}$$ And notice the denominator can be written as $(2k^2-2k+1)(2k^2+2k+1)$

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    $\begingroup$ ...and telescope! $\endgroup$ – sai-kartik Aug 12 '20 at 8:57
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    $\begingroup$ This is a special case of the Sophie-Germain identity where $b = k, a = 1$. $\endgroup$ – Toby Mak Aug 12 '20 at 8:58
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Here's an exhaustive answer: As pointed out by others, the series is not defined well enough. Lagrange interpolation of the first four values gives the cubic $$40x^3-140x^2+200x-95$$ This also passes through the same points as $x^4+1$ does.

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To uniquely determine the sequence $4x^4+1$, the 5th value of the sequence, i.e. $2501$ is required, as that would yield a fourth-degree polynomial.

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As the other answers have already pointed out, once we obtain $T_r = \frac{8r}{4r^4+1}$, solving the problem becomes much easier. $$\begin{gather} S = \sum_{r=1}^{20} \frac{8r}{4r^4+1} \\ = \sum_{r=1}^{20} \frac{2}{2r^2-2r+1} - \frac{2}{2r^2+2r+1} \\ = 2\left(1 - \frac{1}{841}\right) \\ \boxed{S = \frac{1680}{841}} \end{gather}$$

Thank you to everyone who gave their inputs.

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Partial fractions gives: $$A(2k^2 - 2k + 1) + B(2k^2 + 2k + 1) = 8k$$

and substituting $k = 0, 1$ gives $A = 2, B = -2$. Now:

$$f(k)=\frac{2}{2k^2-2k+1} - \frac{2}{2k^2+2k+1}$$

$$f(k+1)=\frac{2}{2k^2+2k+1} - \frac{2}{2(k+1)^2+2(k+1)+1}$$

and the sum telescopes. So $S_{20}$ is:

$$\frac{2}{2(1)^2-2(1)+1} - \frac{2}{2(20)^2 + 2(20) + 1}$$

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