Answer :

^{n}C_{r+1} + ^{n}C_{r-1} + 2 ^{n}C_{r}

= (^{n}C_{r+1} + ^{n}C_{r}) + (^{n}C_{r} + ^{n}C_{r-1})

⇒ (^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r})

= ^{n+1}C_{r+1} + ^{n+1}C_{r}

⇒ (^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r})

= ^{n+2}C_{r+1}

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Prove that the product of 2n consecutive negative integers is divisible by (2n)!

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