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In "New properties for a composition of some generating functions for primes", properties of generating functions are used to form a primality test, speaking specifically in example 1 the following equality appears:

$\displaystyle\sum_{k=1}^{n-1}\binom{n-1}{k-1}\frac{a^{n-k}b^k}{k}=\frac{(a+b)^n-a^n-b^n}{n}, a,b,n\in\mathbb{Z}^+$

I don't doubt its veracity, but I don't see how to justify it (it's probably simple but I'm a bit lost).

Can someone guide me to prove this equality?

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    $\begingroup$ Note that $\binom{n-1}{k-1} = \frac{k}{n}\binom{n}{k}$. $\endgroup$
    – John Omielan
    Commented Jan 19, 2022 at 3:19
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    $\begingroup$ The expression is simply a tricky modification of $(a+b)^n$. $\endgroup$
    – herb steinberg
    Commented Jan 19, 2022 at 3:30

1 Answer 1

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$$\begin{aligned} \sum_{k=1}^{n-1}{n-1\choose k-1}\frac{a^{n-k}b^k}{k}&=\frac{\sum_{k=1}^{n-1}{n \choose k}a^{n-k}b^k}{n} \\ &=\frac{\sum_{k=0}^{n}{n \choose k}a^{n-k}b^k-a^n-b^n}{n} \\ &= \frac{(a+b)^n-a^n-b^n}{n} \end{aligned}$$

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