Question

设数列{an}的前n项和为Sn，对一切n∈N*，点（n,Sn/n）都在函数f(x)=x+an/2x的图像上

令g(x)=(1+2/an)^n(n属于N*),求证2《g(n)<3

Answer 1

点（n,Sn/n）都在函数f(x)=x+an/2x
=》Sn/n=n+an/(2n)
=>Sn=n^2+an/2

an=Sn-S(n-1)=(n^2-(n-1)^2)+(an-a(n-1))/2
=>an=2n-1+an/2-a(n-1)/2
=>an+a(n-1)=4n-2 (a)
=>a(n-1)+a(n-2)=4n-6 (b)
=>an-a(n-2)=4 (i)

又a1=S1=1+a1/2
=>a1=2 (ii)
S2=a1+a2=4*2-2=6
=>a2=4 (iii)
(i）（ii)(iii)=>

=>an=2n

g(n)=(1+2/an)^n=(1+1/n)^n

最后就是归结到证明2<=(1+1/n)^n<3了

Answer 2

这上面不好写。。帮不了你了