歧化砍腿反应
∑ i ( p i ) ( q i ) ( n + i p + q ) = ( n q ) ( n p ) \sum_{i}\binom{p}{i}\binom{q}{i}\binom{n+i}{p+q}=\binom{n}{q}\binom{n}{p} i∑(ip)(iq)(p+qn+i)=(qn)(pn)
∑ i ( p i ) ( q i ) ∑ j ( n j ) ( i p + q − j ) = ∑ i ( q i ) ∑ j ( n j ) ( i p + q − j ) ( p i ) = ∑ i ( q i ) ∑ j ( n j ) ( p p + q − j ) ( j − q i + j − p − q ) = ∑ j ( n j ) ( p p + q − j ) ∑ i ( q i ) ( j − q i + j − p − q ) = ∑ j ( n j ) ( p p + q − j ) ∑ i ( q i ) ( j − q p − i ) = ∑ j ( n j ) ( p p + q − j ) ( j p ) = ∑ j ( n p ) ( n − p j − p ) ( p p + q − j ) = ( n q ) ∑ j ( n − p j − p ) ( p p + q − j ) = ( n q ) ( n q ) \sum_{i}\binom{p}{i}\binom{q}{i}\sum_{j}\binom{n}{j}\binom{i}{p+q-j} \\ =\sum_{i}\binom{q}{i}\sum_{j}\binom{n}{j}\binom{i}{p+q-j}\binom{p}{i} \\ =\sum_{i}\binom{q}{i}\sum_{j}\binom{n}{j}\binom{p}{p+q-j}\binom{j-q}{i+j-p-q} \\ =\sum_{j}\binom{n}{j}\binom{p}{p+q-j}\sum_{i}\binom{q}{i}\binom{j-q}{i+j-p-q} \\ =\sum_{j}\binom{n}{j}\binom{p}{p+q-j}\sum_{i}\binom{q}{i}\binom{j-q}{p-i}\\ =\sum_{j}\binom{n}{j}\binom{p}{p+q-j}\binom{j}{p}\\ =\sum_{j}\binom{n}{p}\binom{n-p}{j-p}\binom{p}{p+q-j}\\ =\binom{n}{q}\sum_{j}\binom{n-p}{j-p}\binom{p}{p+q-j}\\ =\binom{n}{q}\binom{n}{q}\\ i∑(ip)(iq)j∑(jn)(p+q−ji)=i∑(iq)j∑(jn)(p+q−ji)(ip)=i∑(iq)j∑(jn)(p+q−jp)(i+j−p−qj−q)=j∑(jn)(p+q−jp)i∑(iq)(i+j−p−qj−q)=j∑(jn)(p+q−jp)i∑(iq)(p−ij−q)=j∑(jn)(p+q−jp)(pj)=j∑(pn)(j−pn−p)(p+q−jp)=(qn)j∑(j−pn−p)(p+q−jp)=(qn)(qn)
瞎带 p,q 。
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