在数列AN,A1=4,A2=10,若数列LOG3(AN-1),为等差数列,则TN=A1+A1+...+AN-N=?

设bn=log₃(an-1)为等差数列
则b1=log₃(4-1)=log₃3=1
b2=log₃(10-1)=log₃9=2
所以公差为 d=2-1=1
bn=1+(n-1)*1=n
bn=log₃(an-1)=n
an-1=3ⁿ
an=3ⁿ+1
TN=A1+A2+...+AN-N
=3¹+1+3²+1+……+3ⁿ+1-n
=3+3²+3³+……+3ⁿ+n-n
=3*(1-3ⁿ)/(1-3)
=3/2*(3ⁿ-1)