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I'm having a hilariously hard time solving a problem that looks/feels so easy but just won't open up to me. I'm trying to show this equality:

$$\sum_{k=0}^{n-1} \frac{1}{k!(n-k-1)!} = \sum_{k=1}^{n} \frac{1}{(k-1)!(n-k)!}$$

It's so very obvious that the results in these sums are always the same, just that k is shifted by 1 on the right side, but I cannot figure out how extract the right terms to shift the indices properly.

Would really appreciate some help.

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1 Answer 1

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It is useful to make some intermediate passages using another variable. Here we use $j=k+1$; every time that $k$ appears, we write $k=j-1$ instead.

$$ \sum_{k=0}^{(k=)n-1} \frac{1}{k! (n-k-1)!} = \sum_{j-1=0}^{(j-1=)n-1} \frac{1}{(j-1)! (n-(j-1)-1)!} = \sum_{j=1}^{(j=)n} \frac{1}{(j-1)! (n-j+1-1)!} = \sum_{j=1}^{n} \frac{1}{(j-1)! (n-j)!}.$$

After doing this, you change again the variable, going back to $k$ by putting $k=j$:

$$ \sum_{k=0}^{n-1} \frac{1}{k! (n-k-1)!} = \sum_{j=1}^{n} \frac{1}{(j-1)! (n-j)!} = \sum_{k=1}^{n} \frac{1}{(k-1)! (n-k)!} .$$

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  • $\begingroup$ That's really great, thanks! $\endgroup$ Commented Jan 13, 2016 at 1:10

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