www.1862.net > 设等差数列{An}和等比数列{Bn}首项都是1,公差与公...

设等差数列{An}和等比数列{Bn}首项都是1,公差与公...

∵等差数列{an}和等比数列{bn}首项都是1,公差与公比都是2,∴an=1+(n-1)×2=2n-1,bn=2n-1,∴ab1+ab2+ab3+ab4+ab5=a1+a2+a4+a8+a16=1+3+7+15+31=57.故选:D.

由题意可得,an=2n?1,bn=1+(n-1)×1=n由题意可得,在数列{an}中插入的项为,20,1,21,2,3,22,4,5,6,23…2n时,共有项为1+2+…+n+(n+1)=n(1+n)2+n+1=(n+1)(n+2)2当n=62时,63×642=2016即此时共有2016项,且第2016项为262∴c2013=b1951=...

∵{an}是以1为首项,2为公差的等差数列,∴an=2n-1,∵{bn}是以1为首项,2为公比的等比数列,∴bn=2n-1,∴Tn=c1+c2+…+cn=ab1+ab2+…+abn=a1+a2+a4+…+a 2n?1=(2×1-1)+(2×2-1)+(2×4-1)+…+(2×2n-1-1)=2(1+2+4+…+2n-1)-n=2×1?2n1?2-n=2n+1-n-2...

(1)由已知,得an=a+(n-1)b,bn=b?an-1.由a1<b1,b2<a3,得a<b,ab<a+2b.因a,b都为大于1的正整数,故a≥2.又b>a,故b≥3.再由ab<a+2b,得(a-2)b<a.由b>a,故(a-2)b<b,即(a-3)b<0.由b≥3,故a-3<0,解得a<3.于是2≤a<...

∵an=2+(n-1)×1=n+1,bn=2n-1,∴cn=anbn=(n+1)?2n-1,∴Tn=c1+c2+…+cn=2×1+3×2+4×22+5×23+…+(n+1)×2n-1,∴2Tn=2×2+3×22+4×23+…+n×2n-1+(n+1)×2n,∴-Tn=2×2+3×22+4×23+…+n×2n-1+(n+1)×2n=2+(2+22+23+…+2n-1)-(n+1)×2n=2+2(1?2n?1)1?2...

(1)依题意,a5=b5=b1q5-1=1×34=81,故d=a5-a15-1=81-14=20,所以an=1+20(n-1)=20n-19,令Sn=1×1+21×3+41×32+…+(20n-19)?3n-1,①则3Sn=1×3+21×32+…+(20n-39)?3n-1+(20n-19)?3n,②①-②得,-2Sn=1+20×(3+32+…+3n-1)-(20n-19)?3n=1+20×3(1-3n-1)1...

(1)设等差数列{an}的公差为d,∵an=a1+(n-1)d,(a1+d)(a1+13d)=(a1+4d)2(d>0)…(4分)整理:3d2=6a1d(d>0),∴d=2a1=2,∴an=1+(n-1)2=2n-1.∴an=2n-1 (n∈N*)…(7分)(2)bn=2an+1?an+2=2(2n+3)(2n+1)=12n+1-12n+3 …(9分)∴b...

依题得 1+d=q1+5d=q2?d=3q=4,∴an=3n-2,bn=4n-1;∴cn=anbn=(3n-2)?4n-1,∴sn=1?40+4?41+7?42+…+(3n-5)?4n-2+(3n-2)?4n-1,4sn=1?41+4?42+7?43+…+(3n-5)?4n-1+(3n-2)?4n,∴-3sn=1?40+3(41+42+43+…+4n-1)-(3n-2)?4n=1+3×4(1?4n...

∵数列{an}是以1为首项,2为公差的等差数列,∴an=1+2(n-1)=2n-1,∵{bn}是以1为首项,2为公比的等比数列,∴bn=1×2n-1,依题意有:b a1+b a2+…+b a5=b1+b3+b5+b7+b9=1+4+16+64+256=341.故选:D.

(1)∵a2=1+d,a5=1+4d,a14=1+13d∴(1+4d)2=(1+d)(1+13d)∵d>0∴d=2∴an=1+2(n-1)=2n-1∴b2=a2=3,b3=a5=9,故数列{bn}的公比是3,∴bn=3?3n-2=3n-1(2)由c1b1+c2b2+…+cnbn=an+1得当n≥2时,c1b1+c2b2+…+cn?1bn?1=an两式相减得cnbn=an+1-an=...

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