Prove for non-negative integers n,r,k: Σm=0^k(n-m) (r+m)! m! = (r+k+1)! k! [ n r+1 - k r+2 ].

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[JEE Advanced 1991] prove for non negative integers n r k sum m 0 k n m

VAVidaara Admin Asked 2d ago 0 views 1 answer

Prove for non-negative integers $n,r,k$: $\sum_{m=0}^{k}(n-m)\dfrac{(r+m)!}{m!}=\dfrac{(r+k+1)!}{k!}\left[\dfrac n{r+1}-\dfrac k{r+2}\right]$.

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

Answer: Proved.

Split $(n-m)=n-m$, use $\sum_m\binom{r+m}m=\binom{r+k+1}k$ (hockey-stick) and a weighted variant, then simplify to the stated closed form.

JEE Advanced 1991 · Binomial Theorem — verified solution by the Vidaara Team.

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