题目内容：

求证:对任何自然数n,1*2*3...*k+2*3*4...(k+1)+...n(n+1)...(n+k-1)=[n(n+1)...(n+k)]/(k+1)

优质解答

求证: 1*2*3*...*k+2*3*4*...*(k+1)+...+n(n+1)*…*(n+k-1)=[n(n+1)*...*(n+k)]/(k+1) (n为自然数)
证一：数学归纳法.略.
证二：裂项法.
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)
将上面各式求和,得证.
证二的另一描述：
(i+1)(i+2)*...*(i+k)=(-i*(i+1)(i+2)*...*(i+k)+(i+1)(i+2)*...*(i+k)*(i+k+1))/(k+1)
对i=0到n-1累加即证.另外可以用连加号∑(sum)和连乘号∏(prod)来表示,略.
外一则：等效于证
k!/0!+(k+1)!/1!+...+(k+n-1)!/(n-1)!=((k+n)!/n!)/(k+1)
两边同除k!,即证
C(k,0)+C(k+1,1)+...+C(k+n-1,n-1)=C(k+n,n)/(k+1)