設(shè)公比為q,數(shù)列是遞增數(shù)列,q>1
數(shù)列是等比數(shù)列,a1a5=a2a4=729,又a1+a5=246,a1、a5是方程x²-246x+729=0的兩根.
(x-3)(x-243)=0
x=3或x=243
數(shù)列是遞增數(shù)列,a5>a1
a1=3 a5=243
a1/q5=q⁴=81
q>1 q=3
an=a1q^(n-1)=3×3^(n-1)=3ⁿ
數(shù)列{an}的通項公式為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