早教吧作业答案频道 -->数学-->
试证；对任意的正整数n,有1/(1*2*3)+1/(2*3*4)+.+1/n(n+1)(n+2)
题目详情
试证；对任意的正整数n,有1/(1*2*3)+1/(2*3*4)+.+1/n(n+1)(n+2)
▼优质解答
答案和解析
1/n(n+1)(n+2)=1/n·[1/(n+1) - 1/(n+2)]=1/n(n+1) - 1/n(n+2) =[1/n-1/(n+1)] - ½[1/n-1-(n+2)]
=½[1/n-2/(n+1)+1/(n+2)].
∴原式=½(1/1-2/2+1/3)+½(1/2-2/3+1/4)+½(1/3-2/4+1/5)+···+½[1/n-2/(n+1)+1/(n+2)]
=½(1/1+1/2+1/3+···+1/n) - 1·[1/2+1/3+1/4+···+1/(n+1)] + ½[1/3+1/4+···+1/(n+2)]
=½[1+1/2+1/(n+1)+1/(n+2)] + 1·[1/3+1/4+···+1/n] - 1·[1/2+1/3+1/4+···+1/(n+1)]
=1/2+1/4 + ½[1/(n+1)+1/(n+2)] - [1/2+1/(n+1)]
=1/4-½[1/(n+1)-1/(n+2)]
=½[1/n-2/(n+1)+1/(n+2)].
∴原式=½(1/1-2/2+1/3)+½(1/2-2/3+1/4)+½(1/3-2/4+1/5)+···+½[1/n-2/(n+1)+1/(n+2)]
=½(1/1+1/2+1/3+···+1/n) - 1·[1/2+1/3+1/4+···+1/(n+1)] + ½[1/3+1/4+···+1/(n+2)]
=½[1+1/2+1/(n+1)+1/(n+2)] + 1·[1/3+1/4+···+1/n] - 1·[1/2+1/3+1/4+···+1/(n+1)]
=1/2+1/4 + ½[1/(n+1)+1/(n+2)] - [1/2+1/(n+1)]
=1/4-½[1/(n+1)-1/(n+2)]
看了 试证；对任意的正整数n,有1...的网友还看了以下：
已知函数f(x)=x/(2*x+1),数列{an}满足a[1]=1/2,a[n+1]=f(a[n] 2020-05-13 …
设xn=1/1^2+1/2^2+...+1/n^2,证明数列{xn}有极限.请大家知道一二 2020-05-16 …
证明对任意的正整数n,不等式In(n+1)/n<(n+1)/n^2证明对任意的正整数n,不等式In 2020-05-20 …
初等数论的几个问题（1）证明：当n是奇数时,3|2^n+1;当n是偶数时,3不能整除2^n+1（2 2020-06-12 …
若数列{an}中,an不等于0（n>=1),a1=1/2,前n项和Sn满足an=2Sn^2/2Sn 2020-07-09 …
1+1/2+1/3……+1/n,(n>1)证S(2^n)>1+n/2(n>=2,n属于N*）已知S 2020-07-22 …
已知数列{an}得通项公式an=1/n+1+1/n+2+1/n+3+...+1/2n(n∈n*). 2020-07-26 …
利用数学归纳法证明：1*2^1/2*3+2*2^2/3*4+3*2^3/4*5+...+利用数学归 2020-08-01 …
用数学归纳法证明~~~是否存在常数a,b.使1*n+2(n-1)+3(n-2)+...+(n-2) 2020-08-01 …
用数学归纳法证明（n+1)(n+2)…(n+n)=2n·1·3·5·…（2n－1)(n∈N*)时, 2020-08-03 …