已知二項(xiàng)式(2x2-12x)2n的展開式的各項(xiàng)系數(shù)和構(gòu)成數(shù)列{an}.?dāng)?shù)列{bn}的首項(xiàng)b1=1,前n項(xiàng)和為Sn(Sn≠0),且當(dāng)n≥2時,有2S2n=2bnSn-bn(n≥2).
(1)求an和Sn;
(2)設(shè)數(shù)列{(-1)nanSn}的前n項(xiàng)和為Tn,若λ(T2n+19)≤19對任意的正整數(shù)n恒成立,求實(shí)數(shù)λ的取值范圍.
(
2
x
2
-
1
2
x
)
2
n
2
S
2
n
=
2
b
n
S
n
-
b
n
(
n
≥
2
)
{
(
-
1
)
n
a
n
S
n
}
λ
(
T
2
n
+
1
9
)
≤
1
9
【考點(diǎn)】錯位相減法.
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發(fā)布:2024/8/26 6:0:10組卷:40引用:3難度:0.4
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