I want to evaluate the sum : $\sum_{i=0}^n \frac {i^c}{2^i}$ , when c is some positive constant also could you show the way to compute it ? could any one provide way to compute it when c=1 or other value if could not compute the original one !
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The way to compute it is to compute the $n+1$ terms in the sum and add them up. For most $c$, you can't do much better than that, if you want the exact value. For small positive integer values of $c$, it can be done, and has been done on this website, several times (though it's not so easy to find past instances). – Gerry Myerson Sep 17 '13 at 13:27
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I can't find the previous answer. for small natural $c$, we have $\sum_{i=0}^n \frac 1{x^i}=\frac {1-x^{-(n+1)}}{1-\frac 1x}$. Then for $c=1$, $\sum_{i=0}^n \frac i{x^i}=\sum_{i=0}^n- x \frac d{dx}\frac 1{x^i}=- x \frac d{dx}\frac {1-x^{-(n+1)}}{1-x}$, now evaluate it at $x=2$. For larger $c$, repeat the process, applying the operator $-x \frac d{dx} c$ times to each side. I love taking the derivative with respect to $2$.
Ross Millikan
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thank you this exactly what I was looking for ... but the first sum is not correct it should be $\frac{x^{n+1}-1}{x^n(x-1)}$ – Fayez Abdlrazaq Deab Sep 17 '13 at 22:53
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You're right. I was thinking of the sum of $x$, not $\frac 1x$. Fixed. – Ross Millikan Sep 17 '13 at 23:17
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