1. ## binomial theorem questions.

Can anyone please show me the the proof of the ff?

The summation of (nCr)2^r <as r goes from 0 to n> = [3^n + (-1)^n]/2

The summation of r(nCr) <as r goes from 1 to n>= n(2^n-1)
(combinatorial proof)

Your help would be very much appreciated. Thanks.

2. Don't you mean

$\displaystyle \displaystyle \sum_{r=0}^{n} \binom{n}{r}2^r = 3^n$

or

$\displaystyle \displaystyle \sum_{r=1}^{n} \binom{n}{r}2^r = 3^n-1$

These two results follow directly from the application of the binomial theorem in reverse.

$\displaystyle \displaystyle (x+y)^n = \sum_{r=0}^n \binom{n}{r} x^{n-r}y^r.$

with x = 1 and y = 2 where in the second one we are subtracting 1 since we are not summing over the first term which is $\displaystyle 1^n(2^0) = 1$

EDIT: Here is the second one:

$\displaystyle \displaystyle r \binom{n}{r} = r \frac{n!}{(n-r)!r!} = \frac{n!}{(n-r)!(r-1)!} = \frac{n!}{(n-(r-1+1))!(r-1)!} = \frac{n!}{((n-1)-(r-1)))!(r-1)!} =$

$\displaystyle \displaystyle = n\frac{(n-1)!}{((n-1)-(r-1)))!(r-1)!} = n \binom{n-1}{r-1}$

Now the sum becomes:

$\displaystyle \displaystyle \sum_{r=1}^n r \binom{n}{r} = \sum_{r=1}^n n \binom{n-1}{r-1} = n \sum_{r=1}^n \binom{n-1}{r-1} = n \sum_{r=0}^{n-1}\binom{n-1}{r} = n2^{n-1}$