For integers n and r,let binom{n}{r} = begin{ | vedclass

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(A) Using Vandermonde's Identity,$\sum_{i=0}^{k}\binom{n}{i}\binom{m}{k-i} = \binom{n+m}{k}$.Applying this to the given expression:$\sum_{i=0}^{k}\bin

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Not defined

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(A) Using Vandermonde's Identity,$\sum_{i=0}^{k}\binom{n}{i}\binom{m}{k-i} = \binom{n+m}{k}$.Applying this to the given expression:$\sum_{i=0}^{k}\binom{10}{i}\binom{15}{k-i} = \binom{10+15}{k} = \binom{25}{k}$.Similarly,$\sum_{i=0}^{k+1}\binom{12}{i}\binom{13}{k+1-i} = \binom{12+13}{k+1} = \binom{25}{k+1}$.The total sum is $\binom{25}{k} + \binom{25}{k+1}$.Using Pascal's Identity,$\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$,we get $\binom{25}{k} + \binom{25}{k+1} = \binom{26}{k+1}$.Since $\binom{n}{r}$ is defined for all non-negative integers $n$ and $r$ (with the given condition),the expression $\binom{26}{k+1}$ is defined for any non-negative integer $k$. Therefore,there is no maximum value for $k$.

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