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)
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/22 13:53:45
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)
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)
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)
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)
数学归纳法是一种方法
也可以直接用裂项求和的方法
n(n+1)...(n+k-1) = n(n+1)...(n+k-1)[(n+k)-(n-1)]/(k+1)
= [(n-1)n(n+1)…(n+k-1) - n(n+1)…(n+k-1)(n+k)]/(k+1)
然后求和就可以了
用数学归纳法吧