Prove the identity Σk=0^m 2n-k k 2n-k n · 2n-4k+1 2n-2k+1 2^ n-2k= nm 2n-2m n-m 2^ n-2m.

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[JEE Advanced 1999] prove the identity sum k 0 m binom 2n k k binom 2n k

VAVidaara Admin Asked 2d ago 0 views 1 answer

#PYQJEE #JEE #JEE Advanced #1999 #binomial-sums #Binomial Theorem

Prove the identity $\sum_{k=0}^{m}\dfrac{\binom{2n-k}k}{\binom{2n-k}n}\cdot\dfrac{2n-4k+1}{2n-2k+1}\,2^{\,n-2k}=\dfrac{\binom nm}{\binom{2n-2m}{n-m}}\,2^{\,n-2m}$.

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

Answer: Proved.

Telescoping: the $k$-th term equals $f(k)-f(k+1)$ for a suitable $f$, and the partial sum collapses to the stated right-hand side.

JEE Advanced 1999 · Binomial Theorem — verified solution by the Vidaara Team.

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