求證: 1*2*3*...*k+2*3*4*...*(k+1)+...+n(n+1)*…*(n+k-1)=[n(n+1)*...*(n+k)]/(k+1) (n為自然數(shù))
證一：數(shù)學歸納法.略.
證二：裂項法.
1*2*3...*k= (-0*1*2*3...*k+1*2*3...*k*(k+1))/(k+1)
2*3...*k*(k+1)= (-1*2*3...*k*(k+1)+2*3...*(k+1)*(k+2))/(k+1)
...
n(n+1)*…*(n+k-1)=(-(n-1)n(n+1)*…*(n+k-1)+n(n+1)*...*(n+k)]/(k+1)
將上面各式求和,得證.
證二的另一描述：
(i+1)(i+2)*...*(i+k)=(-i*(i+1)(i+2)*...*(i+k)+(i+1)(i+2)*...*(i+k)*(i+k+1))/(k+1)
對i=0到n-1累加即證.另外可以用連加號∑(sum)和連乘號∏(prod)來表示,略.
外一則：等效于證
k!/0!+(k+1)!/1!+...+(k+n-1)!/(n-1)!=((k+n)!/n!)/(k+1)
兩邊同除k!,即證
C(k,0)+C(k+1,1)+...+C(k+n-1,n-1)=C(k+n,n)/(k+1)