数列（高手进）我有数列的题不会A设Sn是等差数列{an}的前n项和,若a5/a3=5/9,则S9/S5的值为多少B数列1,1/1+2,1/1+2+3,...,1/1+2+3+...+n的前十项和为多少

Answer 1

A.S9/S5=[(a1+a9)*9/2]/[(a1+a5)*5/2]
=(9/5)*(a1+a9)/(a1+a5)
=(9/5)*(2*a5)/2*a3)
=(9/5)*(5/9)
=1
B.题目中应该为1,1/(1+2).
第n项=1/(1+2+3+...+n)=2/[n(n+1)]=2[1/n-1/(n+1)].
所以前n项和
Sn=2(1/1-1/2)+2(1/2-1/3)+2(1/3-1/4)+...+2[1/n-1/(n+1)]
=2[1/1-1/(n+1)]
=2n/(n+1).
所以前十项和=20/11