问题 解答题

已知等差数列{an}和公比为q(q≠1)的正项等比数列{bn}满足a1=b1=a,a3=b3,a7=b5

(1)求等比数列{bn}的公比q;

(2)记Mn=a1+a2+…+an,Nn=b1+b2+…+bn,试比较M5与N5的大小.

(3)若a=1,设数列cn=a2n+1•b2n+1,求数列{cn}的前n项和Sn

答案

(1)因为a1=b1,a3=b3,a7=b5

所以,a1+2d=b1q2=aq2
a1+6d=b1q4=aq4

变形得:a(1-q2)=-2d ①

a(1-q4)=-6d ②

②÷①得,1+q2=3

正项等比数列{bn},
所以q2=2,

即,q=

2

(2)由(1)可知d=

a
2

M5=

5×(a+a+4d)
2
=10a;

N5=

a(1-(
2
)
5
)
1-
2
=
a((
2
)
5
-1)
2
-1
=
a(2
2
-1)
2
-1

M5
N5
=
10a 
a(2
2
-1)
2
-1
=
10(
2
-1)
2
2
-1
>1,

M5>N5

(3)an=a+(n-1)

a
2
=
a
2
(n+1)
bn=a•(
2
)
n-1

由题意若a=1,数列cn=a2n+1•b2n+1=(2n+2)•(

2
)2n=(n+1)•2n+1

Sn=2•22+3•23+4•24+…+(n+1)•2n+1…①

2Sn=2•23+3•24+4•25+…+(n+1)•2n+2…②

②-①得Sn=-2•22-3•23-24-…-2n+1+(n+1)•2n+2

Sn=-4+(n+1)•2n+2-

4(1-2n-1)
1-2
=-4+(n+1)•2n+2+4(1-2n-1)=(n+1)•2n+2-2n+1

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