已知{an}是各项均为正数的等比数列,且a1+a2=2(
(Ⅰ)求{an}的通项公式; (Ⅱ)设bn=an2+log2an,求数列{bn}的前n项和Tn. |
(Ⅰ)设等比数列{an}的公比为q,则an=a1qn-1,由已知得:
,a1+a1q=2(
+1 a1
) 1 a1q a1q2+a1q3 =32(
+1 a1q2
) 1 a1q3
化简得:
,即a12q(q+1)=2(q+1) a12q5 (q+1)=32(q+1)
,a12q=2 a12q5=32
又a1>0,q>0,解得:
,a1=1 q=2
∴an=2n-1;
(Ⅱ)由(Ⅰ)知bn=an2+log2an=4n-1+(n-1)
∴Tn=(1+4+42+…+4n-1)+(1+2+3+…+n-1)
=
+4n-1 4-1 n(n-1) 2
=
+4n-1 3
.n(n-1) 2