Compute $\sum_{n=1}^{1010} (-1)^{n+1} \binom{2020-n}{n-1} 2^{2021-2n}.$

Question

Compute $$\sum_{n=1}^{1010} (-1)^{n+1} \binom{2020-n}{n-1} 2^{2021-2n}.$$

Everything is in the summation

I wanted to test out the first few values but then $2^{2021-2n}$ gets too large.

One thing i noticed was that i could take out the 1 and -1 on odd/even terms respectively and make the summation slightly easier. Something like:$$\sum_{n=1}^{505} \binom{2021-2n}{2n-2} 2^{2023-4n} - \sum_{n=1}^{505} \binom{2020-2n}{2n-1} 2^{2021-4n}$$

would this be the way to go or is this bad? and if yes, how would i continue

Answer 1

Let’s let $k=n-1$ and rewrite the summation as

$$\sum_{k=0}^{1009}(-1)^k\binom{2019-k}k2^{2019-2k}\,.$$

The binomial coefficient in the last term is $\binom{1010}{1009}$; had we increased $k$ any further, we’d have had $\binom{1009}{1010}=0$. Thus, we might as well write the sum as

$$\sum_{k\ge 0}(-1)^k\binom{2019-k}k2^{2019-2k}\,.$$

This is the case $n=2019$ of

$$f(n)=\sum_{k\ge 0}(-1)^k\binom{n-k}k2^{n-2k}\,,$$

so it’s reasonable to compute $f(n)$ for some small values of $n$. We quickly find that $f(0)=1$, $f(1)=2$, $f(3)=4$, $f(4)=5$, and $f(5)=6$, which suggests that we may have an identity

$$f(n)=\sum_{k\ge 0}(-1)^k\binom{n-k}k2^{n-2k}=n+1\,.$$

This is quite easily proved by induction on $n$, and for now I’ll leave that to you. Of course it yields an answer of $f(2019)=2020$ to your specific question.