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](/uploads/image/z/13318532-44-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D3%2C3a%EF%BC%88n%2B1%EF%BC%89%3Dan%28n%3D1%2C2%2C3..%29%2C%E8%AE%BEbn%3Dan%2Blog%EF%BC%883%EF%BC%89an%EF%BC%88n%3D1%2C2%2C3..%EF%BC%89%E5%88%99%7Bbn%7D%E7%9A%84%E5%89%8D%E6%95%B0%E5%88%97%E5%92%8CSn)
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