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Find a formula for \(\sum_{i=0}^\infty i^{2014}x^{i}\). In other words, find \(0^{2014}x^{0}+1^{2014}x^{1}+2^{2014}x^{2}+3^{2014}x^{3}+4^{2014}x^{4}+...\).

Hint: Try a form of recursion(not with numbers, but with a formula).

Have fun!

Find a formula for \(\sum_{i=0}^\infty i^{2014}x^{i}\). In other words, find \(0^{2014}x^{0}+1^{2014}x^{1}+2^{2014}x^{2}+3^{2014}x^{3}+4^{2014}x^{4}+...\).

Hint: Try a form of recursion(not with numbers, but with a formula).

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## Comments

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TopNewestWe can get a recurrence relation in the following manner:

For positive integers k we define \(T_{k}(x)=\sum_{i=0}^{\infty }i^kx^i\) \(\forall x \in (0,1)\)

We know \(T_{1}(x)=\frac{x}{(1-x)^2}\)

Also \(T_{k+1}(x)=x\frac{dT_{k}(x)}{dx}\) – Sambit Senapati · 2 years, 9 months ago

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– Tristan Shin · 2 years, 9 months ago

To be precise, is there a way of expanding your recursion formula?Log in to reply

\(T_{k+1}(x)=T_{k}(x)+x^2T_{k-1}''(x)\) – Sambit Senapati · 2 years, 9 months ago

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For a start, I labeled \(T_{k}=\frac {f(x)}{(1-x)^{k+1}}\). Then, try working around with that. – Tristan Shin · 2 years, 9 months ago

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Just a note, I know how to do it, but it would take a very long time(probably over a month straight) of working out some functions. I'm not really looking for an exact solution(don't actually take the time to do it), but if anyone can discover the method I'm using, that would be wonderful. – Tristan Shin · 2 years, 9 months ago

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