已知数列{an}满足a1=3,3a（n+1）=an(n=1,2,3..),设bn=an+log（3）an（n=1,2,3..）则{bn}的前数列和Sn

显示全部楼层 · 2013-1-15 16:31:37

3a[n+1]=ana[n+1]/an=1/3故有an=a1q^(n-1)=3*(1/3)^(n-1)=3^(2-n)bn=an+log3(an)=3^(2-n)+2-n=(1/3)^(n-2)+(2-n)=9*(1/3)^n+(2-n)Sn=9*1/3*(1-1/3^n)/(1-1/3)+(1+2-n)n/2=9/2*(1-1/3^n)+(3-n)n/2...

千问 · 2013-1-15 16:31:37

1 3a(n+1)=ana(n+1)/an=1/3q=1/3an=3*(1/3)^(n-1)=3^(2-n)bn=3^(2-n)+log3 3^(2-n)=3^(2-n)+2-nSn=3+1+……+3^(2-n)+(1+0+……+2-n)={3*(1/3)^n/[1-(1/3)]}+[1+(2-n)]*n/2={[9*(1/3)^n]/2}+...

千问 · 2013-1-15 16:31:37

3a(n+1)=ana(n+1)/an = 1/3an/a1= (1/3)^(n-1)an = 3^(-n)bn= an + log(3)an
= 3^(-n) - nSn =b1+b2+...+bn
= (1-3^(-n) )/ (1-1/3)- n(n+1)/2
= (3/2)(1-...

千问 · 2013-1-15 16:31:37

两个等比数列，很好计算，楼上已经解出...