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)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...

an=3+(n-1)=n+2,bn=12n,∴anbn=(n+2)?12n,∴Sn=3?12+4?122+…+(n+2)?12n,∴12Sn=3?122+4?123+…+(n+1)?12n+(n+2)?12n+1,两式相减,化简可得Sn=4-n+42n.故答案为:4-n+42n.

(1)由题意得(1+4d)2=(1+d)(1+13d),d>0解得d=2…(3分)∴an=2n-1…(4分)又b2=a2=3,b3=a5=9,所以{bn}的公比为3,bn=3n-1…(6分)(2)∵cn=2an-18=4n-20…(7分)令cn≤0得n≤5…(9分)所以当n=4或n=5时,sn取最小值-40…(12分)

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...

设等比数列{an}的公比为q>0,∵a1=2,a3=a2+4,∴2q2=2q+4,化为q2-q-2=0,∵q>0,解得q=2.∴an=2×2n?1=2n.∵{bn}是首项为1,公差为2的等差数列,∴bn=1+2(n-1)=2n-1.∴数列{an+bn}的前n项和Sn=a1+a2+…+an+b1+b2+…+bn=21+22+…+2n+(1+3+…+2n-1...

∵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)设{an}的公差为d=2,{bn}的公比为q=2,∵a1=b1=1∴an=1+2(n-1)=2n-1,bn=qn?1=2n?1.(2)∵anbn=(2n?1)?2n?1,∴Sn=1×20+3×21+5×22+…+(2n?1)?2n?1,①2Sn=,1×21+3×22+…+(2n?3)?2n?1+(2n?1)?2n,②相减得:?Sn=1+2(21+22+…+2n?1)?(2n?1)...

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