# Thread: Prove by using Binomial Theorem

1. ## Prove by using Binomial Theorem

$\displaystyle \sum i \binom{n}{i}=n2^{n-1}$

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.

2. Originally Posted by thelaughingman
$\displaystyle \sum_{i=1}^n i \binom{n}{i}=n2^{n-1}$
Are you also allowed to use a bit of calculus? If so, write down the binomial expansion of $\displaystyle (x+1)^n$, differentiate it and then put $\displaystyle x=1$.

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

4. Using the binomial theorem with $\displaystyle a = b = 1$, putting $\displaystyle m = n-1$ then multiplying both sides by $\displaystyle n$, we have:

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