Prove this .

**thereddevils** · February 18th 2009, 12:24 AM

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

**CaptainBlack** · February 18th 2009, 12:32 AM

Originally Posted by thereddevils

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

Telescoping series:

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

CB

**thereddevils** · February 18th 2009, 12:38 AM

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

**CaptainBlack** · February 18th 2009, 12:48 AM

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 $0\times 2^0=0$ is a give away for writting the general term as $r2^r-(r-1)2^{r-1}$ , since then the only part of the last term of the series that does not cancel is $n2^2$ , and from the first term is $0$ .

CB

**thereddevils** · February 18th 2009, 04:03 AM

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

$<br /> (r+1)2^{r-1}=r2^r-(r-1)2^{r-1}<br />$

**mr fantastic** · February 18th 2009, 04:09 AM

Originally Posted by thereddevils

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

$<br /> (r+1)2^{r-1}=r2^r-(r-1)2^{r-1}<br />$

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

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