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,由已知条件得当 n大于1时 a1+a2/2+a3/3+a4/4……+an/n-a1+a2/2+a3/3+a4/4……+an-1/n-1= n^2-2n-2-[（n-1）*（ n-1）-2（ n-1）-2]= 2n -3= an/n,则an= 2n^2-3n ,当n =1时,a1=1^2-2*1-2=-3不等于2*1^2-3*1=-1,所以 1 an=2n^2-3n,n大于1时 ；2 an= 2n^2-3n ,当n =1时.2题 由已知条件得an=2n,又Sn=na1+n（n-1）K/2,K为公差,可以得Sn=n*n+n,则1/Sn=1/n*n+n=1/n*n+n,
1/s1+1/s2+1/s4……1/sn=（1-1/2+1/2-1/3+1/3……+1/n-1/n+1）=n/n+1