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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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    I'm doubtful about the validity of the equation for $n=0$, but that's a minor nitpick.2012-04-14

2 Answers 2

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$1=\frac{b}{x}+\frac{a}{y}$. 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$2012-04-15
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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?2012-04-14