数列难题组1.设数列{an}满足：a1+a2/2+a3/3+a4/4……+aan/n=n^2-2n-2求数列{an}的通项公式2.已知等差数列前三项为2,4,6前n项和为Sn,S50=2550求1/s1+1/s2+1/s4……1/sn=?

Question

数列难题组
1.设数列{an}满足：a1+a2/2+a3/3+a4/4……+aan/n=n^2-2n-2 求数列{an}的通项公式
2.已知等差数列前三项为2,4,6前n项和为Sn,S50=2550
求1/s1+1/s2+1/s4……1/sn=?

Answer 1

1.设an/n=bn,bn的前N项和为Sn
Sn=n2-2n-2
Sn-1=n2-4n+1
所以bn=Sn-Sn-1=2n-3=an/n
所以an=n(2n-3)
2.a1=2,公差d=2,则an=2n
Sn=(2+2n)*n/2=n(n+1)
所以1/Sn=1/[n(n+1)]=1/n - 1/(n+1)
所以,所求=1-1/2+1/2-1/3+……+1/n-1/(n+1)=1-1/(n+1)=n/(n+1)