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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.

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    $\begingroup$ 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 $(1+x)^{2n}$ $\endgroup$ – saulspatz Oct 5 '18 at 14:55
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Hint:

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

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

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  • $\begingroup$ The idea is excellent, but the use of $a,b$ is rather misleading. Do you need them? $\endgroup$ – user376343 Oct 5 '18 at 17:46
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\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 $$ (\bullet)=2018\cdot 2017 \cdot 2^{2015}=1009 \cdot 2017 \cdot 2^{2016} $$

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