Question

已知数列an的前n项和为Sn，若满足Sn=2an+（-1）^n

Answer 1

1)an=Sn-S(n-1)=2an+(-1)^n-2a(n-1)-(-1)^(n-1)=2an-2a(n-1)+2*(-1)^n
得:an=2a(n-1)-2*(-1)^n
化为:an+(2/3)(-1)^n=2[a(n-1)+(2/3)(-1)^(n-1)]
因此{an+(2/3)(-1)^n}为公比是2的等比数列
a1=S1=2a1-1,得a1=1
a1+(2/3)(-1)=1/3
因此有an+(2/3)(-1)^n=(1/3)*2^(n-1)
得:an=(1/3)*2^(n-1)-(2/3)(-1)^n
2) m>4时,am>0
左边=3[1/(2^3-2)+1/(2^4+2)+1/(2^5-2)+....]
1/(2^4+2)+1/(2^5-2)=(2^5+2^4)/[(2^4+2)(2^5-2)]=(2^5+2^4)/[2^9+2^6-2^5-4]<(2^5+2^4)/2^9=1/2^4+1/2^5
1/(2^6+2)+1/(2^7-2)=(2^7+2^6)/[(2^6+2)(2^7-2)]=(2^7+2^6)/[2^13+2^8-2^7-4]<(2^7+2^6)/2^13=1/2^6+1/2^7
....
所以左边<3[1/6+1/2^4+1/2^5+1/2^6+1/2^7+....]
<3(1/6+1/8)
=7/8