已知數(shù)列{an}為正項(xiàng)等比數(shù)列,數(shù)列{bn}滿足b1=1,b2=3,a1b1+a2b2+a3b3+?+anbn=3+(2n-3)2n.
(1)求an;
(2)設(shè){bnan}的前n項(xiàng)和為Sn,證明:Sn<6.
a
1
b
1
+
a
2
b
2
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a
3
b
3
+
?
+
a
n
b
n
=
3
+
(
2
n
-
3
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2
n
{
b
n
a
n
}
【考點(diǎn)】錯(cuò)位相減法.
【答案】(1);
(2)證明見解析.
a
n
=
2
n
-
1
(2)證明見解析.
【解答】
【點(diǎn)評】
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發(fā)布:2024/6/27 8:0:9組卷:36引用:2難度:0.6
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