For 2 le r le n,the expression binom{n}{r} + | 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 the middle term $2\binom{n}{r-1}$ as $\binom{n}{r-1} + \b

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The given expression is $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$.We can rewrite the middle term $2\binom{n}{r-1}$ 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 $2 \le r \le n$,the expression $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$ is equal to:

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