# Prove by using Binomial Theorem

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• September 19th 2010, 01:08 PM
thelaughingman
Prove by using Binomial Theorem
$\sum i \binom{n}{i}=n2^{n-1}$

The summation is from i=0 to n but I couldnt figure out how to put that in with latex.

I tried expanding out the Left Side but that didn't lead me anywhere.

I thought about splitting the right side's 2 into a form of (x+y)^n but I couldn't figure out what my x and y should be.

Help please?
• September 19th 2010, 01:34 PM
Opalg
Quote:

Originally Posted by thelaughingman
$\sum_{i=1}^n i \binom{n}{i}=n2^{n-1}$

Are you also allowed to use a bit of calculus? If so, write down the binomial expansion of $(x+1)^n$, differentiate it and then put $x=1$.
• September 19th 2010, 01:51 PM
thelaughingman
Quote:

Originally Posted by Opalg
Are you also allowed to use a bit of calculus? If so, write down the binomial expansion of $(x+1)^n$, differentiate it and then put $x=1$.

Damn it, I thought of splitting 2 into (1+1) but never thought of (x+1) and subbing x=1 later.

I'm not sure if this is a valid solution but I'll find out tomorrow.

Thanks a lot!
• September 20th 2010, 03:33 AM
TheCoffeeMachine
Using the binomial theorem with $a = b = 1$, putting $m = n-1$ then multiplying both sides by $n$, we have:

$\displaystyle 2^m = \sum_{k=0}^{m}\binom{m}{k} \Rightarrow \displaystyle 2^{n-1} = \sum_{k=0}^{n-1}\binom{n-1}{k} \Rightarrow \displaystyle n2^{n-1} = \sum_{k=0}^{n-1}{n}\binom{n-1}{k} = \sum_{k=0}^{n}{k}\binom{n}{k}.$