For 2 le r le n,binom{n}{r} + 2binom{n}{r-1} | vedclass

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(D) The given expression is $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$.We can rewrite $2\binom{n}{r-1}$ as $\binom{n}{r-1} + \binom{n}{r-1}$.So

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$\binom{n+2}{r}$

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(D) The given expression is $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$.We can rewrite $2\binom{n}{r-1}$ as $\binom{n}{r-1} + \binom{n}{r-1}$.So,the expression becomes $\binom{n}{r} + \binom{n}{r-1} + \binom{n}{r-1} + \binom{n}{r-2}$.Using the Pascal's identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$,we get:$(\binom{n}{r} + \binom{n}{r-1}) + (\binom{n}{r-1} + \binom{n}{r-2}) = \binom{n+1}{r} + \binom{n+1}{r-1}$.Applying the identity again,$\binom{n+1}{r} + \binom{n+1}{r-1} = \binom{n+2}{r}$.Thus,the correct option is $D$.

For $2 \le r \le n$,$\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$ is equal to

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