已知正项等比数列{an}是递增数列,且满足a1+a5=246,a2a4=729期(1)求数列an的通项公式(2)设bn=anlog3an+1(n∈N*),数列bn的前n项和为Tn,求Tn
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![已知正项等比数列{an}是递增数列,且满足a1+a5=246,a2a4=729期(1)求数列an的通项公式(2)设bn=anlog3an+1(n∈N*),数列bn的前n项和为Tn,求Tn](/uploads/image/z/400698-18-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%AD%A3%E9%A1%B9%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E6%98%AF%E9%80%92%E5%A2%9E%E6%95%B0%E5%88%97%2C%E4%B8%94%E6%BB%A1%E8%B6%B3a1%2Ba5%3D246%2Ca2a4%3D729%E6%9C%9F%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E8%AE%BEbn%3Danlog3an%2B1%EF%BC%88n%E2%88%88N%2A%EF%BC%89%2C%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E6%B1%82Tn)
已知正项等比数列{an}是递增数列,且满足a1+a5=246,a2a4=729期(1)求数列an的通项公式(2)设bn=anlog3an+1(n∈N*),数列bn的前n项和为Tn,求Tn
已知正项等比数列{an}是递增数列,且满足a1+a5=246,a2a4=729期(1)求数列an的通项公式
(2)设bn=anlog3an+1(n∈N*),数列bn的前n项和为Tn,求Tn
已知正项等比数列{an}是递增数列,且满足a1+a5=246,a2a4=729期(1)求数列an的通项公式(2)设bn=anlog3an+1(n∈N*),数列bn的前n项和为Tn,求Tn
设公比为q,数列是递增数列,q>1
数列是等比数列,a1a5=a2a4=729,又a1+a5=246,a1、a5是方程x²-246x+729=0的两根.
(x-3)(x-243)=0
x=3或x=243
数列是递增数列,a5>a1
a1=3 a5=243
a1/q5=q⁴=81
q>1 q=3
an=a1q^(n-1)=3×3^(n-1)=3ⁿ
数列{an}的通项公式为an=3ⁿ
bn=anlog3[a(n+1)]=(n+1)×3ⁿ
Tn=b1+b2+...+bn=2×3+3×3²+4×3³+...+(n+1)×3ⁿ
3Tn==2×3²+3×3³...+n×3ⁿ+(n+1)×3^(n+1)
Tn-3Tn=-2Tn=2×3+3²+3³+...+3ⁿ-(n+1)×3^(n+1)
=1+3+...+3ⁿ -(n+1)×3^(n+1) +2
=1×[3^(n+1)-1]/(3-1) -(n+1)×3^(n+1) +2
=-(2n+1)×3^(n+1)/2 + 3/2
Tn=(2n+1)×3^(n+1) /4 -3/4