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I'm doing some induction practice from the textbook Problems on Algorithms, and I couldn't figure out this problem:

For even $n \ge 4$ and $ 2 \le i \le \frac{n}{2}$: $$ \sum_{k=1}^{i} \prod_{j=1}^{k} (n-2j+1) \le 2\prod_{j=1}^i(n-2j+1) $$

I wrote out a couple of the terms and I got the following:

$$ (n-2+1) + (n-2+1)(n-4+1) + (n-2+1)(n-4+1)(n-6+1)+... \le 2[(n-2+1)(n-4+1)(n-6+1)...] $$

I'm not sure how to prove that the IH holds for $n+2$ if it holds for $n$ (I got the base case and the trivial parts). How should I approach this?

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  • $\begingroup$ There seems to be something wrong. The case $n=6,i=3$ gives $5+5\cdot 3+5\cdot 3\cdot 1\le 2\cdot 5\cdot 3\cdot 1$ which is not true. $\endgroup$ Commented Nov 8, 2017 at 20:34
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    $\begingroup$ You're right.. it seems that the inequality should be the other way. I guess it's a typo in the book. If $i=\frac{n}{2}$, then the LHS is greater than or equal to the RHS for all cases $\endgroup$
    – Peter Wang
    Commented Nov 8, 2017 at 20:44

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