已知數(shù)列?an?的前n項(xiàng)和為Sn,且a1=4..(Ⅰ)求數(shù)列?an?的通項(xiàng)公式,(Ⅱ)設(shè)數(shù)列?bn?滿足:b1=4.且bn+1=bn2-(n-1)bn-2(n∈N*),求證:bn>an(n≥2.n∈N*), 查看更多

 

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已知數(shù)列{an}的前n項(xiàng)和為Sn,且a1=4,Sn=nan+2-
n(n-1)
2
,(n≥2,n∈N*)

(Ⅰ)求數(shù)列{an}的通項(xiàng)公式;
(Ⅱ)設(shè)數(shù)列{bn}滿足:b1=4,且bn+1=bn2-(n-1)bn-2,(n∈N*),
求證:bn>an,(n≥2,n∈N*);
(Ⅲ)求證:(1+
1
b2b3
)(1+
1
b3b4
)(1+
1
b4b5
)…(1+
1
bnbn+1
)<
3e

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已知數(shù)列{an}的前n項(xiàng)和為Sn,且a1=4,Sn=nan+2-
n(n-1)
2
,(n≥2,n∈N*)

(I)求數(shù)列{an}的通項(xiàng)公式;
(II) 已知bn>an,(n≥2,n∈N*),求證:(1+
1
b2b3
)(1+
1
b3b4
)(1+
1
b4b5
)…(1+
1
bnbn+1
3e

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已知數(shù)列{an}的前n項(xiàng)和為Sn,且a1=4,Sn=nan+2-
n(n-1)2
,(n≥2,n∈N*
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)設(shè)數(shù)列{bn}滿足:b1=4,且bn+1=bn2-(n-1)bn-2(n∈N*),求證:bn>an,(n≥2,n∈N*).

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已知數(shù)列{an}的前n項(xiàng)和為Sn點(diǎn)Pn(an,-
1
an+1
)
(n∈N*)在曲線f(x)=-
4+
1
x2
上,且a1=1,an>0.
(1)求證:數(shù)列{
1
a
2
n
}
是等差數(shù)列,并求an;
(2)數(shù)列{bn}的前n項(xiàng)和為Tn,且滿足
Tn+1
a
2
n
=
Tn
a
2
n+1
+16n2-8n-3
,設(shè)定b1的值,使得數(shù)列{bn}是等差數(shù)列;
(3)求證:Sn
1
2
4n+1
-1
(n∈N*).

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已知數(shù)列{an}的前n項(xiàng)和為Sn,
(1)若點(diǎn)(n,Sn)均在函數(shù)y=f(x)的圖象上,且f(x)=3x2-2x,求{an}的通項(xiàng)公式;
(2)若a1=a2=1,且
an+1
an
an
an-1
(0<λ<1,n=2,3,4…),證明:
a1+k
a1
+
a2+k
a2
+…+
an+k
an
λk
1-λk
(常數(shù)k∈N*且k≥3)

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