1*2*3+2*3*4+...+n(n+1)(n+2)=在数列{an}中,an=1/(n+1)+2/(n+1)+...+n/(n+1).又bn=2/(an*an-1) 求数列{bn}的n项和sn=1/2+1/6+...+1/n(n+1) 若sn*sn+1=3/4

1*2*3+2*3*4+...+n(n+1)(n+2)=在数列{an}中,an=1/(n+1)+2/(n+1)+...+n/(n+1).又bn=2/(an*an-1) 求数列{bn}的n项和sn=1/2+1/6+...+1/n(n+1) 若sn*sn+1=3/4
1*2*3+2*3*4+...+n(n+1)(n+2)=
在数列{an}中,an=1/(n+1)+2/(n+1)+...+n/(n+1).又bn=2/(an*an-1) 求数列{bn}的n项和
sn=1/2+1/6+...+1/n(n+1) 若sn*sn+1=3/4

1*2*3+2*3*4+...+n(n+1)(n+2)=在数列{an}中,an=1/(n+1)+2/(n+1)+...+n/(n+1).又bn=2/(an*an-1) 求数列{bn}的n项和sn=1/2+1/6+...+1/n(n+1) 若sn*sn+1=3/4
n(n+1)(n+2)=n^3+3n^2+2n
1^3+2^3+...+n^3=[n(n+1)/2]^2（自然数立方和公式,可以用更高次的项来推,这里省略）;
n项自然数平方和公式：n(n+1)(2n+1)/6;
则1*2*3+2*3*4+...+n(n+1)(n+2)=[n(n+1)/2]^2+n(n+1)(2n+1)/6+n(n+1)
这里还可以提取n(n+1),留给你了
2
先求出an=1/(n+1)+2/(n+1)+...+n/(n+1)=n/2
然后裂项相消
n>=2时
bn=2/(an*an-1)=8/[n(n-1)]=8[1/(n-1)-1/n]
则前n项和为8(n-1)/n;
3
1/n(n+1) =1/n-1/(n+1)
则sn=1-1/2+1/2-1/3+...+1/n-1/(n+1)=n/(n+1)
又sn*sn+1=n/(n+1)*(n+1)/(n+2)=n/(n+2)=3/4
解得n=6