已知a1=9,点(an,an+1)在函数f(x)=x2+2x的图象上,其中n∈N*,设bn=lg(1+an).(Ⅰ)证明数列{bn}是等比数列;(Ⅱ)设cn=nbn,求数列{cn}的前n项和Sn;(Ⅲ

题目简介

已知a1=9,点(an,an+1)在函数f(x)=x2+2x的图象上,其中n∈N*,设bn=lg(1+an).(Ⅰ)证明数列{bn}是等比数列;(Ⅱ)设cn=nbn,求数列{cn}的前n项和Sn;(Ⅲ

题目详情

已知a1=9,点(an,an+1)在函数f(x)=x2+2x的图象上,其中n∈N*,设bn=lg(1+an).
(Ⅰ) 证明数列{bn}是等比数列;
(Ⅱ) 设cn=nbn,求数列{cn}的前n项和Sn
(Ⅲ) 设dn=
1
an
+
1
an+2
,求数列{dn}的前n项和Dn
题型:解答题难度:中档来源:不详

答案

(Ⅰ) 证明:由题意知:an+1=
a2n
+2an

an+1+1=(an+1)2
∵a1=9∴an+1>0,
lg(an+1+1)=lg(an+1)2,即bn+1=2bn.
又∵b1=lg(1+a1)=1>0,
∴{bn}是公比为2的等比数列.
(Ⅱ) 由(1)知:bn=b12n-1=2n-1,∴cn=n•2n-1
∴Sn=c1+c2+…+cn=1•20+2•21+3•22+…+n•2n-1①,
2Sn=1•21+2•22+3•23+…+(n-1)•2n-1+n•2n②,
∴①-②得,-Sn=1•20+21+22+…+2n-1-n•2n=
1-2n
1-2
-n•2n=2n-1-n•2n

S n=n•2n-2n+1
(Ⅲ)∵an+1=
a2n
+2an=an(
a n
+2)>0

class="stub"1
an+1
=class="stub"1
2
(class="stub"1
an
-class="stub"1
an+2
)
,∴class="stub"1
an+2
=class="stub"1
an
-class="stub"2
an+1

dn=class="stub"1
an
+class="stub"1
an
-class="stub"2
an+1
=2(class="stub"1
an
-class="stub"1
an+1
)

Dn=d1+d2+…+dn=2(class="stub"1
a1
-class="stub"1
a2
+class="stub"1
a2
-class="stub"1
a3
+…class="stub"1
an
-class="stub"1
an+1
)=2(class="stub"1
a1
-class="stub"1
an+1
)

又由(1)知:lg(1+an)=2n-1
an+1=102n-1,∴an+1=102n-1
Dn=2(class="stub"1
9
-class="stub"1
102 n-1
)

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