# Prove this .

• February 18th 2009, 01:24 AM
thereddevils
Prove this .
Prove that $\sum^{n}_{r=1}(r+1)2^{r-1}=n2^n$ .
• February 18th 2009, 01:32 AM
CaptainBlack
Quote:

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
• February 18th 2009, 01:38 AM
thereddevils
Thanks Captain , but how did you get that expression ? Is it by expansion ?
• February 18th 2009, 01:48 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 $0\times 2^0=0$ 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
• February 18th 2009, 05:03 AM
thereddevils
I know what a telescoping series is but i just don understand how you get this :

$
(r+1)2^{r-1}=r2^r-(r-1)2^{r-1}
$
• February 18th 2009, 05:09 AM
mr fantastic
Quote:

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

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

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