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If $x-y=z$, show that : $$1=(-1)^{m+1}\sum_{j=0}^{m} \dbinom{m+j}{j}\left[\frac{x^jy^{m+1}}{z^{m+j+1}}-(-1)^{m+j+1}\frac{x^{m+1}y^j}{z^{m+j+1}}\right]$$

I tried expanding: $$1=\left(\frac{x}{z}-\frac{y}{z}\right)^{m+1}$$ as it was given as a hint in the book but couldn't reach far enough ..

Dexter
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2 Answers2

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We have two sums (here $x-y=z$)

$$A = (-1)^{m+1} \frac{y^{m+1}}{z^{m+1}} \sum_{j=0}^m {m+j\choose j} \frac{x^j}{z^j}$$

and

$$B = \frac{x^{m+1}}{z^{m+1}} \sum_{j=0}^m {m+j\choose j} (-1)^j \frac{y^j}{z^j}$$

and we want to show that

$$A+B = 1.$$

Dividing by $z$ we find

$$\frac{x}{z}-\frac{y}{z} = 1.$$

We may put $x/z = p$ and $-y/z = 1-p.$ We get for $A$ and $B$

$$A = (1-p)^{m+1} \sum_{j=0}^m {m+j\choose j} p^j$$

and

$$B = p^{m+1} \sum_{j=0}^m {m+j\choose j} (1-p)^j.$$

This is now revealed to be an instance of an identity by Gosper, which appeared at this MSE link.

Marko Riedel
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For $x=2,y=1,z=1\ and\ m=1$, we have: $$RHS = (-1)^2 \cdot \sum_{j=0}^{2} \dbinom{j+1}{j} \left[x^j - (-1)^{j+2}x^2 \right] $$ $$= \sum_{j=0}^{2}(j+1)\left[2^j - (-1)^j\cdot4\right]$$ $$= 1\cdot (1-4)\ +\ 2\cdot(2+4)\ +\ 3\cdot(4-4)$$ $$=9$$ Therefore I think this is not true.

Deepesh Thakur
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