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.
(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)由(1)可知d=,
M5==10a;
N5===,
==>1,
M5>N5.
(3)an=a+(n-1)=(n+1),bn=a•()n-1
由题意若a=1,数列cn=a2n+1•b2n+1=(2n+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+(n+1)•2n+2+4(1-2n-1)=(n+1)•2n+2-2n+1.