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? ​