Question

已知数列an的前n项和为sn,且sn=4/3an-1/3

设数列{an}的前n项的和为Sn,且Sn=4/3an-1/3乘以2^(n+1)+2/3(n属于N,n≥1）
(1)证明:数列{an+2^n}是等比数列,并求通项an（2）设bn=（2^n）/Sn,n=1,2,3,……,证明n∑i=1 bi小于3/2（n在∑上面,i=1在下面）

Answer 1

1.Sn=4/3an-1/3*2^(n+1)+2/3,Sn+1=4/3an+1-1/3*2^(n+2)+2/3,S1=a1=4/3a1-1/3*4+2/3=2
Sn+1-Sn=an+1=4/3an+1-4/3an-1/3*2^(n+2)+1/3*2^(n+1)
化简得到4/3an+4/3*2^n=1/3an+1+1/3*2^(n+1)
(an+1+2^(n+1))/ (an+2^n)=4,所以{an+2^n}是等比数列且第一项为4,通项为4^n
所以an=4^n-2^n,Sn=4(1-4^n)/(1-4)+2(1-2^n)
2.设tn=2^n
则bn=(3/4)tn/(tn^2-3/2tn+1/2)=(3/2)tn/((tn-1)(2tn-1))=(3/2)(1/(tn-1)-1/(2tn-1))
将tn带入=(3/2)(1/(2^n-1)-1/(2^(n+1)-1))
提出3/2后可发现当中项可以全部抵消就留下头和尾
=3/2(1-1(2^(n+1)-1))
应为1(2^(n+1)-1))当n趋向无穷大时该值趋向0+所以n∑i=1 bi小于3/2