Binomial theorem/Pascal's identity help

[Moistwh]

Hi I have no idea how to evaluate this sum

I know I either have to change the sum or change the binomial coefficient, but Pascal's identity only alters it by 1, not by i.

 

[LCKurtz]

Hi I have no idea how to evaluate this sum
View attachment 13868
I know I either have to change the sum or change the binomial coefficient, but Pascal's identity only alters it by 1, not by i.

Hint: What do you get if you expand [MATH](1 + \frac 1 3)^n[/MATH]?

 

[pka]

Hi I have no idea how to evaluate this sum
View attachment 13868
I know I either have to change the sum or change the binomial coefficient, but Pascal's identity only alters it by 1, not by i.

\(\displaystyle \sum\limits_{k = 0}^n \binom{n}{k}3^{-k}=\left(1+\dfrac{1}{3}\right)^n\)

 

[Moistwh]

Omg I just realised n choose i is the same thing as n choose (n-i).

Thanks for the help