∑i=0n(a+ia)⋅(b+n−ib)=(a+b+n+1n)\sum_{i=0}^n\binom{a+i}a\cdot\binom{b+n-i}b=\binom{a+b+n+1}n∑i=0n(aa+i)⋅(bb+n−i)=(na+b+n+1)
