Question

已知等差数列｛an}的前n项通项公式为Sn，且a1+a3=10,S4=24 令Tn==1/S1+……1/Sn,求证Tn＜3/4

Answer 1

{an}的公差为d
a1+a3=a1+a1+2d=10
得到：a1+d=5 ①
S4=4(a1+a4)/2=2(a1+a1+3d)=24
得到：2a1+3d=12 ②
综合①② 得到：a1=3 d=2

Sn=na1+n(n-1)d/2=n(n+2)
1/Sn=1/n(n+2)=½[1/n-1/(n+2)] 这是关键
Tn=1/S1+……1/Sn
Tn=½[1/1-1/3+1/2-1/4+1/3-1/5+。。。+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=½[1+1/2-1/(n+1)-1/(n+2)]
＜3/4

Answer 2

a1+a3=10,a1+d=5
S4=24 2a1+3d=12
构成方程组，得a1=3,d=2
Sn=3n+n(n-1)=n^2+2n
1/sn=1/[n(n+2)] =1/2(1/n-1/(n+2)]
Tn=1/2[（1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+……+(1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1) -1/(n+2) ]<1/2(1+1/2)=3/4
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