Use Binomial theorem to simplify $\sum_{k=0}^{16}(-1)^k\binom{16}k (3)^{16-k}$

Question

When I simplified it out to equal $(x + y)^{n}$ which is $(3 - 1)^{16}$ or 65,536. Is that correct or have I messed up somewhere?

I hope my question even comes out right, I did my best with the mathjax but I can't see the output of my title (n on top of the sum is 16, but I couldnt get a 2 digit number to fit on top)

$$\sum_{k=0}^{16}(-1)^k\binom{16}k (3)^{16-k}$$

Answer 1

Recall the binomial theorem:

$$(a+b)^n = \sum_{k=0}^n \binom n k a^k b^{n-k}$$

Take $a = -1, b=3,n=16$: then

$$(-1+3)^{16} = \sum_{k=0}^{16} \binom {16} k (-1)^k 3^{16-k}$$

which is identical to your sum and simplifies to $2^{16}$ as claimed.

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In short, yes, you're correct. I just hope that writing it as I have above can resolve any doubts. If absolutely nothing else, Wolfram Alpha also agrees with us.