Induction Proof with combination

[srfriggen]

Homework Statement

Prove:

$\sum$$^{n}_{r=0}$2^r($^{n}_{r}$) = 3ⁿ

Homework Equations

The Attempt at a Solution

I proceeded by induction:

Testing the base case for n=0 is correct.


Moving right along to try to show:

$\sum$$^{n+1}_{r=0}$2^r($^{n}_{r}$) = 3ⁿ⁺¹



This is where I'm getting stuck. I can obtain:

3²+2ⁿ⁺¹($^{n+1}_{n}$)

Which I think equals: 3²+2ⁿ⁺¹($^{n}_{n-1}$)

Which equals: 3²+2ⁿ⁺¹*n




I am not sure how to proceed after this. Any help would be greatly appreciated.

also, this was a problem on a test I took yesterday.

 

[tiny-tim]

hi srfriggen!

much easier would be a proof using the binomial expansion ((a + b)ⁿ)

 

[srfriggen]

So I can say, by the binomial theorem:

3ⁿ=(2+1)ⁿ=$\sum$$^{n}_{r}$($^{n}_{r}$)2^r*1^n-r

 

[tiny-tim]

yup!

quicker? ​