Question

数列{a_n}的前n项和记为S_n,a₁=t,点(S_n,a_n+1)在直线y=3x+1上,n∈N^*.
(1)当实数t为何值时,数列{a_n}是等比数列?
(2)在(1)的结论下,设b_n=log₄a_n+1,c_n=a_n+b_n,T_n是数列{c_n}的前n项和,求T_n.

Answer 1

(1) t=1    (2) T_n=+

解析解:(1)∵点(S_n,a_n+1)在直线y=3x+1上,
∴a_n+1=3S_n+1,a_n=3S_n-1+1(n>1),
a_n+1-a_n=3(S_n-S_n-1)=3a_n,
∴a_n+1=4a_n,
a₂=3S₁+1=3a₁+1=3t+1,
∴当t=1时,a₂=4a₁,数列{a_n}是等比数列.
(2)在(1)的结论下,a_n+1=4a_n,a_n+1=4ⁿ,
b_n=log₄a_n+1=n,
c_n=a_n+b_n=4^n-1+n,
T_n=c₁+c₂+…+c_n
=(4⁰+1)+(4¹+2)+…+(4^n-1+n)
=(1+4+4²+…+4^n-1)+(1+2+3+…+n)
=+.