# Prove this .

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• Feb 18th 2009, 12:24 AM
thereddevils
Prove this .
Prove that $\displaystyle \sum^{n}_{r=1}(r+1)2^{r-1}=n2^n$ .
• Feb 18th 2009, 12:32 AM
CaptainBlack
Quote:

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

Telescoping series:

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

CB
• Feb 18th 2009, 12:38 AM
thereddevils
Thanks Captain , but how did you get that expression ? Is it by expansion ?
• Feb 18th 2009, 12:48 AM
CaptainBlack
Quote:

Originally Posted by thereddevils
Thanks Captain , but how did you get that expression ? Is it by expansion ?

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 AM
thereddevils
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 AM
mr fantastic
Quote:

Originally Posted by thereddevils
I know what a telescoping series is but i just don understand how you get this :

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

$\displaystyle r 2^{r-1} + 2^{r-1} = r 2^{r-1} (2 - 1) + 2^{r-1} = r 2^r - r 2^{r - 1} + 2^{r-1} = r 2^r - 2^{r-1} (r - 1)$.