Prove that $\displaystyle \sum^{n}_{r=1}(r+1)2^{r-1}=n2^n$ .

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- Feb 18th 2009, 12:24 AMthereddevilsProve this .
Prove that $\displaystyle \sum^{n}_{r=1}(r+1)2^{r-1}=n2^n$ .

- Feb 18th 2009, 12:32 AMCaptainBlack
- Feb 18th 2009, 12:38 AMthereddevils
Thanks Captain , but how did you get that expression ? Is it by expansion ?

- Feb 18th 2009, 12:48 AMCaptainBlack
First I deduce what method is appropriate for high school level, which made me suspect a telescoping series, then noting that $\displaystyle 0\times 2^0=0$ is a give away for writting the general term as $\displaystyle r2^r-(r-1)2^{r-1}$, since then the only part of the last term of the series that does not cancel is $\displaystyle n2^2$, and from the first term is $\displaystyle 0$.

CB - Feb 18th 2009, 04:03 AMthereddevils
I know what a telescoping series is but i just don understand how you get this :

$\displaystyle

(r+1)2^{r-1}=r2^r-(r-1)2^{r-1}

$ - Feb 18th 2009, 04:09 AMmr fantastic