已知数列an的前n项和Sn,且an>0,n属于N*,an,Sn,an^2成等差数列

发布网友 发布时间：2024-10-24 17:46

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热心网友 时间：2024-10-28 22:34

an + (an)^2 = 2Sn
n=1
(a1)^2 -a1=0
a1= 1

an + (an)^2 = 2Sn
Sn = (1/2) {an + (an)^2}
an = Sn -S(n-1)
=(1/2) {an + (an)^2} - (1/2) {a(n-1) + [a(n-1)]^2}
[ (an)^2 - an ] -[a(n-1))^2 + a(n-1) ]=0
{(an)^2 - [a(n-1)]^2} - [an+a(n-1)]=0
[an+a(n-1)][ an- a(n-1) -1 ] =0
an- a(n-1) -1 =0
an- a(n-1) =1
an -a1 = n-1
an =n
Sn = n(n+1)/2
1/Sn = 2[1/n -1/(n+1)]
Tn =1/S1+1/S2 +...+1/Sn
= 2[ 1 - 1/(n+1) ]
= 2n/(n+1)