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$\sum_{i=1}^n i2^i$

How can we derive a function out of this summation. I saw similar type of equation before but the $\lvert r\rvert$ in $ir^i$ is < 1. How can we create a function when $\lvert r\rvert > 1$ (in this case $2$)

Cup
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  • In the title of the question you asked about $2^n$ instead of $2^i$. Which of both is correct? I assume $2^i$ because $2^n$ would be trivial. – Martin Rosenau Oct 21 '18 at 04:55
  • The accepted answer to the question you linked to gives a formula for $S_{m}=\sum_{n=1}^{m}nr^{n}$ which is valid for any $r \ne 1$. It's only the final step (taking the limit) which requires $|r|<1$. – Misha Lavrov Oct 21 '18 at 04:56
  • @MishaLavrov From the answer I linked, how did he get $S_{m} - rS_{m} = -mr^{m+1}+\sum_{n=1}^{m}r^{n}$ – Cup Oct 21 '18 at 05:00
  • If you want to get rid of the $\sum$: I doubt that this is possible. Look at $r=10$ and $n<10$; in this case the results are numbers like 43210 or 876543210. I doubt that there is a possibility to write such numbers without the use of the $\sum$ sign. – Martin Rosenau Oct 21 '18 at 05:03
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    Write $r S_m = \sum_{n=1}^{m+1}(n-1)r^n $ and subtract the two sums term-by-term. – Misha Lavrov Oct 21 '18 at 05:03
  • http://www.ebyte.it/library/docs/math06b/PowerSeriesKpXk.html – user514787 Oct 21 '18 at 05:08

1 Answers1

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Using differentiation. From the well known identity $$1+x+x^2+x^3+...+x^{n-1}+x^{n}=\frac{x^{n+1}-1}{x-1} \tag{1}$$ we have $$\sum\limits_{i=1}^{n}x^i=\frac{x^{n+1}-1}{x-1}-1=\frac{x^{n+1}-x}{x-1}$$ Now take the derivative $$\left(\sum\limits_{i=1}^{n}x^i\right)^{'}=\left(\frac{x^{n+1}-x}{x-1}\right)^{'}$$ which is $$\sum\limits_{i=1}^{n}ix^{i-1}=\frac{n x^{n+1}-(1+n)x^n+1}{(x-1)^2}$$ and multiply both sides by $x$ $$\sum\limits_{i=1}^{n}ix^{i}=\frac{nx^{n+2}-(1+n)x^{n+1}+x}{(x-1)^2}\tag{2}$$ and set $x=2$ $$\sum\limits_{i=1}^{n}i2^{i}=n2^{n+2}-(1+n)2^{n+1}+2\tag{3}$$

rtybase
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