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

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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

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
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

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
3a[n+1]=an
a[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

1 3a(n+1)=an
a(n+1)/an=1/3
q=1/3
an=3*(1/3)^(n-1)=3^(2-n)
bn=3^(2-n)+log3 3^(2-n)
=3^(2-n)+2-n
Sn=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}+[n(3-n)/2]
=[3^(2-n)/2]+[n(3-n)/2]

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

两个等比数列,很好计算,楼上已经解出

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