Evaluate $\sum_{k=0}^{50}{{100}\choose{2k}}18^{2k}$

Question

I am trying to evaluate the following sum: $$\sum_{k=0}^{50}{{100}\choose{2k}}18^{2k}$$

I thought about using the binomial theorem, but it doesn't exactly fit. How can I approach something like that?

Answer 1

You have that $$19^{100}=(18+1)^{100}=\sum_{k=0}^{100} {100 \choose k} 18^k$$ and $$17^{100}= \sum_{k=0}^{100} (-1)^k {100 \choose k} 18^k$$ Now, adding side by side you obtain $$19^{100}+17^{100}=2 \sum_{k=0}^{50} {100 \choose 2k} 18^{2k}$$ So, your answer is $\frac{19^{100}+17^{100}}2$.

Answer 2

By the binomial Theorem we have $$\sum_{k=0}^{n} \binom{2n}{2k} x^k=\frac12\left((\sqrt x-1)^{2n}+(\sqrt x+1)^{2n}\right)$$

Answer 3

Hint: think about $(18+1)^{100}$ and $(18-1)^{100}$