# Summation involving factorials

The question at hand is:

$$\sum_{n=1}^{1008} \ \frac{2018!}{(2018-2n)!(2n-2)!} = a * 2^{b}$$

where $$a, b > 0$$ and GCD of $$(a,2)$$ = $$1$$

Find the remainder when $$a$$ is divided by 1025.

My approach: Evaluating the first few terms, I got:

$$\frac{2018!}{2016!0!} + \frac{2018!}{2014!2!} + \frac{2018!}{2012!4!} ..... + \frac{2018!}{2!2014!}$$

I know that it's easier to simplify symmetric sums as a part of permuations and combinations but this sum doesn't seem to be symmetric as it's missing the last $$\frac{2018!}{0!2016!}$$ term.

I also noticed that the terms follow a pattern like: $$\binom{2016}{2x}$$ as your $$x$$ goes from $$1$$ to $$1008$$, but I can't seem to get a sense on how to apply that here.

• Substitute $x=1$ in the binomial formula for $(1-x)^{2n}$ and then compare to the result of substituting $x=1$ in the formula for $(1+x)^{2n}$ – saulspatz Oct 5 '18 at 14:55

Hint:

$$(a+b)^{2m}+(a-b)^{2m}=?$$

Put $$a=b=1,2m=2016$$

• The idea is excellent, but the use of $a,b$ is rather misleading. Do you need them? – user376343 Oct 5 '18 at 17:46

\begin{align*} (\bullet)&=\sum_{n=1}^{1008} \frac{2018!}{(2018-2n)!(2n-2)!}=2018\cdot 2017 \sum_{n=1}^{1008} \frac{2016!}{(2016-2n+2)!(2n-2)!}=\\ &= 2018\cdot 2017 \sum_{n=1}^{1008} \binom{2016}{2n-2}=2018\cdot 2017 \sum_{n=0}^{1007} \binom{2016}{2n} \end{align*} Claim: $$\sum_{n=0}^{p-1}\binom{2p}{2n}=2^{2p-1}$$ Proof. Use that \begin{align*} 2^p &=\sum_{n=0}^{p}\binom{p}{n}, \space\ \space\ \sum_{n}^{k}(-1)^n \binom{p}{n}=(-1)^k \binom{p-1}{k} \end{align*}

\begin{align*} 2\sum_{n=0}^{p-1}\binom{2p}{2n} &=2\left(\sum_{n=0}^{p}\binom{2p}{2n}-1 \right)=\sum_{n=0}^{2p}\binom{2p}{n}+\sum_{n=0}^{2p}(-1)^n\binom{2p}{n}=2^{2m}+1+\sum_{n=0}^{2p-1}(-1)^n\binom{2p}{n}=\\ &=2^{2p}+1+(-1)^{2p-1}\binom{2p-1}{2p-1}=2^{2p} \end{align*}

Thus $$(\bullet)=2018\cdot 2017 \cdot 2^{2015}=1009 \cdot 2017 \cdot 2^{2016}$$