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A friend of mine asked me to help with this problem. I tried induction, but I didn't know how to get this formula.

If $x$ and $y$ are real numbers such that $xy= ax+by$. Show that $$ x^ny^n=\sum_{k=1}^{n}{2n-1-k \choose n-1}(a^n b^{n-k}x^k+ a^{n-k}b^n y^k), \forall n \geq 0 $$

Any help is appreciated.

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How is this related to computer science? –  Antonio Vargas Apr 14 '12 at 18:34
It is a question from "Concrete Mathematics, A Foundation for Computer Science". –  yaa09d Apr 14 '12 at 18:42
I removed the (computer-science) tag. –  user2468 Apr 14 '12 at 18:43
I'm doubtful about the validity of the equation for $n=0$, but that's a minor nitpick. –  hardmath Apr 14 '12 at 19:30
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2 Answers

$1=\frac{b}{x}+\frac{a}{y}$. So, let $s=\frac{b}{x}, t=\frac{a}{y}$. It is enough to show that $$2=\sum_{k=-\infty}^n \binom{2n-1-k}{n-1}( s^n t^{n-k}+s^{n-k} t^n).$$ Now, let $m=n-k$. Then $$ RHS= (st)^n \sum_{m=0}^{\infty} \binom{n-1+m}{m} (s^{m-n}+t^{m-n})=2.$$ Note that $$(1-z)^{-n}=\sum_{m=0}^{\infty} \binom{n-1+m}{m} z^m.$$

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Thanks. But Why is the LHS of the first identity is $2$, and why does the sum start from $-\infty$ –  yaa09d Apr 15 '12 at 5:54
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Hint : Replace $(xy)^n$ (LaTeX notation) on the left by $(ax+by)^n$ expand the left-hand side using the Newton formula, and replace products $x^i y^j$ by $(ax+by)^{min(i,j)},$ etc. Nice exercice !

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Thanks for the hint. But how can I shorten the expression after that? –  yaa09d Apr 14 '12 at 19:53
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