已知数列an满足：an>0,且对一切n属于N*,有a1^3+a2^3+…+an^3=Sn^2,其中Sn为数列an的前n项和.（1）求数列an的通项公式（2）证明1/lna2+1/lna3+…+1/lnan>3n^2-n-2/2n(n+1)(n大于等于2,n属于N*)第二问如何解答

Question

已知数列an满足：an>0,且对一切n属于N*,有a1^3+a2^3+…+an^3=Sn^2,其中Sn为数列an的前n项和.
（1）求数列an的通项公式
（2）证明1/lna2+1/lna3+…+1/lnan>3n^2-n-2/2n(n+1)(n大于等于2,n属于N*)
第二问如何解答

Answer 1

（1）当n=1时,S1^2=a1^3=a1^2,因为a1>0,所以a1=1
当n≥2时,S(n-1)^2=a1^3+a2^3+…+a(n-1)^3,则Sn^2-S(n-1)^2=an^3,即[Sn+S(n-1)][Sn-S(n-1)]=an^3,而Sn-S(n-1)=an①,且an>0,所以Sn+S(n-1)=an^2②.①+②得：2Sn=an^2+an
所以2S(n-1)=a(n-1)^2+a(n-1),则2Sn-2S(n-1)=an^2+an-[a(n-1)^2+a(n-1)]=2an,所以有an^2-an-a(n-1)^2-a(n-1)=0,化简得：[an-a(n-1)-1][an+a(n-1)]=0,因为an>0,所以an+a(n-1)>0,所以an-a(n-1)-1=0,即an-a(n-1)=1,所以数列an是以1为首项、1为公差的等差数列,则an=n (n∈N+)
（2）令f(x)=(x^2-1)/2-lnx (x≥2),则f'(x)=x-1/x=(x+1)(x-1)/x,当x≥2时f'(x)>0,所以f(x)在[2,+∞)上单调递增,则f(x)≥f(2)=1.5-ln2>0,所以(x^2-1)/2-lnx>0,则(x^2-1)/2>lnx,则(n^2-1)/2>lnn (n≥2)
所以1/lnn>2/(n^2-1)=[1/(n-1)]-[1/(n+1)]
所以1/lna2+1/lna3+…+1/lnan=1/ln2+1/ln3+…+1/lnn
>1-1/3+1/2-1/4+1/3-1/5+…+1/[1/(n-2)]-(1/n)+[1/(n-1)]-[1/(n+1)]
=1+1/2-1/n-1/(n+1)
=(3n^2-n-2)/2n(n+1),得证.