Using the Binomial Theorem to Show 3^k C(n,k) = 3^k 1^n-k C(n,k)

[jenny Downer]

how do i use binomial to show that 3^k C(n,k) = 3^k 1^n-k C(n,k)

 

[CRGreathouse]

Did you type that correctly? Right now you have stuff = stuff * 1, which is clearly correct...

 

[jenny Downer]

i wanted to explain a bit more

 

[jenny Downer]

It should be
n
sum 3^k C(n,k) = 2^2n
k=0

the hint is 3^k C(n,k) = 3^k 1^n-k C(n,k)

 

[HallsofIvy]

The reason C(n, k) are called "binomial coefficients" is that C(n, k) is the coefficient of x^k in (x+ y)ⁿ What are the coefficients of x^k in (3x+ y)ⁿ? What do you get if x= y= 1?