If for 3 leq r leq 30, binom{30}{30-r} + 3bin | vedclass

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(C) Recall the Pascal's identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$.The given expression is $\binom{30}{30-r} + 3\binom{30}{31-r} + 3\bi

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(C) Recall the Pascal's identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$.The given expression is $\binom{30}{30-r} + 3\binom{30}{31-r} + 3\binom{30}{32-r} + \binom{30}{33-r}$.We can rewrite the coefficients $3$ as $1+2$ to group terms:$= \binom{30}{30-r} + \binom{30}{31-r} + 2\binom{30}{31-r} + 2\binom{30}{32-r} + \binom{30}{32-r} + \binom{30}{33-r}$$= [\binom{30}{30-r} + \binom{30}{31-r}] + 2[\binom{30}{31-r} + \binom{30}{32-r}] + [\binom{30}{32-r} + \binom{30}{33-r}]$Using Pascal's identity, this simplifies to:$= \binom{31}{31-r} + 2\binom{31}{32-r} + \binom{31}{33-r}$$= [\binom{31}{31-r} + \binom{31}{32-r}] + [\binom{31}{32-r} + \binom{31}{33-r}]$$= \binom{32}{32-r} + \binom{32}{33-r} = \binom{33}{33-r}$.By the symmetry property $\binom{n}{k} = \binom{n}{n-k}$, we have $\binom{33}{33-r} = \binom{33}{33-(33-r)} = \binom{33}{r}$.Comparing this with $\binom{m}{r}$, we get $m = 33$.

If for $3 \leq r \leq 30$, $\binom{30}{30-r} + 3\binom{30}{31-r} + 3\binom{30}{32-r} + \binom{30}{33-r} = \binom{m}{r}$, then $m$ equals:

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