f(x)=x2-x+b(b∈R)
∵y=f(x)的图象过原点,
∴f(x)=x2-x,
∴Sn=n2-n
∴an=Sn-Sn-1
=n2-n-[(n-1)2-(n-1)]
=2n-2(n≥2)
∵a1=S1=0
所以,数列{an}的通项公式为
an=2n-2(n∈N*)
(2)由an+log3n=log3bn得:
bn=n•32n-2(n∈N*)
Tn=b1+b2+b3++bn
=30+2•32+3•34++n•32n-2(1)
∴9Tn=30+2•32+3•34++n•32n(2)
(2)-(1)得:8Tn=n•32n−(30+32+34++32n−2)=n•32n−
32n−1 |
8 |
∴Tn=
n•32n |
8 |
32n−1 |
64 |
(8n−1)32n+1+1 |
64 |