已知正項(xiàng)數(shù)列{an}中,a1=3,2Sn+2Sn-1=a2n-3(n≥2).
(1)求{an}的通項(xiàng)公式;
(2)若bn=an2n,求{bn}的前n項(xiàng)和Tn.
a
1
=
3
,
2
S
n
+
2
S
n
-
1
=
a
2
n
-
3
(
n
≥
2
)
b
n
=
a
n
2
n
【考點(diǎn)】錯位相減法.
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【解答】
【點(diǎn)評】
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發(fā)布:2024/6/27 10:35:59組卷:110引用:7難度:0.5
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