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JEE MainMathematicsModerateBinomial Theorem

The sum of
series  50C050!50!+ 50C149!51!+ 50C248!52!\frac{\ ^{50}C_{0}}{50!50!} + \frac{\ ^{50}C_{1}}{49!51!} + \frac{\ ^{50}C_{2}}{48!52!}+.........+ 50C50100!\frac{\ ^{50}C_{50}}{100!} equal
to

Show answer and solution
∑r=050\mspace2mu 50Cr(50−r)!(50+r)!\sum_{r = 0}^{50}\mspace{2mu}\frac{\ ^{50}C_{r}}{(50 - r)!(50 + r)!}
=1100!∑r=050\mspace2mu( 50Cr 100C50−r)= \frac{1}{100!}\sum_{r = 0}^{50}\mspace{2mu}\left( \ ^{50}C_{r}\ ^{100}C_{50 - r} \right)
=1100!( 50C0 100C50+ 50C1 100C49+………+ 50C50 100C0)= \frac{1}{100!}\left( \ ^{50}C_{0}\ ^{100}C_{50} + \ ^{50}C_{1}\ ^{100}C_{49} + \ldots\ldots\ldots + \ ^{50}C_{50}\ ^{100}C_{0} \right)
=1100!( 150C50)= \frac{1}{100!}\left( \ ^{150}C_{50} \right)

Source: Grass-G / M-18 Mathematics Paper 27 Nov Third.docx

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