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If $$^nJ_r = \frac{(1-x^n)(1-x^{n-1})(1-x^{n-2})\cdots (1-x^{n-r+1})}{(1-x)(1-x^2)(1-x^3)\cdots (1-x^r)}$$

then prove that $\displaystyle ^nJ_{n-r}=^nJ_{r}$

what i try

$$^nJ_{n-r} = \frac{(1-x^n)(1-x^{n-1})(1-x^{n-2})\cdots (1-x^{r+1})}{(1-x)(1-x^2)(1-x^3)\cdots (1-x^{n-r})}$$

dod not kow how do i solve it help me please

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  • $\begingroup$ Set them equal and clear denominators. You get a true statement. Now just do the steps in reverse. $\endgroup$ – saulspatz Feb 11 at 14:27
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    $\begingroup$ why not try some concrete examples, like $n=10, r=3$, to see what's going on? $\endgroup$ – Matthew Towers Feb 11 at 14:29
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    $\begingroup$ It might help to write out explicitly a (slightly) non-trivial example, say ${}^7 J_3$ and see where the cancellations come from. $\endgroup$ – NickD Feb 11 at 14:30
  • $\begingroup$ did not understand fully please explain me $\endgroup$ – jacky Feb 11 at 14:48
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Let $\displaystyle f(r)=(1-x)(1-x^2)(1-x^3)\cdots\cdots (1-x^r).$

Then $\displaystyle ^nJ_{r}=\frac{(1-x^n)(1-x^{n-1})\cdots (1-x^{n-r+1})\cdots (1-x^{n-r})(1-x^{n-r-1})\cdots (1-x)}{(1-x)(1-x^2)\cdots (1-x^{n-r})}\cdots \times \frac{1}{(1-x)(1-x^2)\cdots (1-x^{n-r})}.$

$$\displaystyle ^nJ_{r}=\frac{f(n)}{f(r)\cdot f(n-r)}.$$

$$ ^nJ_{r}=^nJ_{n-r}.$$

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