已知数列{an}的前n项和为Sn,且满足Sn=2an-2n(n∈N*)?(I)设bn=an+2,求数列{bn}的通项公式;(II)若
已知数列{an}的前n项和为Sn,且满足Sn=2an-2n(n∈N*)?(I)设bn=an+2,求数列{bn}的通项公式;(II)若数列{cn}满足cnlog2bn,求数...
已知数列{an}的前n项和为Sn,且满足Sn=2an-2n(n∈N*)?(I)设bn=an+2,求数列{bn}的通项公式;(II)若数列{cn}满足cnlog2bn,求数列{cnbn}的前n项和Tn.
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(Ⅰ)∵Sn=2an-2n(n∈N*),
∴当n≥2时,Sn-1=2an-1-2(n-1).
两式相减得an=2an-2an-1-2,即an=2an-1+2(n≥2).…(3分)
又∵a1=2,可知an>0,
∴当n≥2时,
=
=
=2(常数),
∴{bn}是以b1=a1+2=4为首项,2为公比的等比数列,
∴数列{bn}的通项公式bn=2n+1.…(6分)
(Ⅱ)∵cn=log2bn=log22n+1=n+1,
∴
=
,…(8分)
则Tn=
+
+…+
+
,…①
Tn=
+
+…+
+
,…②
两式相减得,
Tn=
+
+
+…+
?
…(10分)
=
+
∴当n≥2时,Sn-1=2an-1-2(n-1).
两式相减得an=2an-2an-1-2,即an=2an-1+2(n≥2).…(3分)
又∵a1=2,可知an>0,
∴当n≥2时,
bn |
bn?1 |
an+2 |
an?1+2 |
2an?1+4 |
an?1+2 |
∴{bn}是以b1=a1+2=4为首项,2为公比的等比数列,
∴数列{bn}的通项公式bn=2n+1.…(6分)
(Ⅱ)∵cn=log2bn=log22n+1=n+1,
∴
cn |
bn |
n+1 |
2n+1 |
则Tn=
2 |
22 |
3 |
23 |
n |
2n |
n+1 |
2n+1 |
1 |
2 |
2 |
23 |
3 |
24 |
n |
2n+1 |
n+1 |
2n+2 |
两式相减得,
1 |
2 |
2 |
22 |
1 |
23 |
1 |
24 |
1 |
2n+1 |
n+1 |
2n+2 |
=
1 |
4 |
|