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设等差数列{An}和等比数列{Bn}首项都是1,公差与公...

(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}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...

∵a1<b1,b2<a3,∴a<b以及ba<a+2b∴b(a-2)<a<b,a-2<1?a<3,a=2.又因为 am+3=bn?a+(m-1)b+3=b?an-1.又∵a=2,b(m-1)+5=b?2n-1,则b(2n-1-m+1)=5.又b≥3,由数的整除性,得b是5的约数.故2n-1-m+1=1,b=5,∴an=a+b(n-1)=2+5(n-...

(1)∵a1<b1<a2<b2<a3∴a<b<a+b<ab<a+2b.∵ab>b,a,b都为正整数,∴a>1∵ab<a+2b,∴(a-2)b<a.∵b>a,∴(a-2)b<b,即(a-3)b<0.∵b为正整数,∴a-3<0,解得a<3.∵a∈N,∴a=2; (2)由(1)知a=2,则am=2+(m-1)b,bn=b?22n-1...

(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<...

(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...

∵数列{an}是以d为公差的等差数列,且a1=d,∴a2=2d,a3=3d.a12+a22+a32=14d2.又数列{bn}是公比q的等比数列,且b1=d2,∴b2=d2q,b3=d2q2.∴a12+a22+a32b1+b2+b3=14d2d2(1+q+q2)=141+q+q2∈N*.∵q是正整数,∴1+q+q2=7,解得q=2.∴S92T8=(9d+9...

(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=...

(Ⅰ)∵a<b<a+b<ab<a+2b,a,b∈N*,∴a+b<abab<a+2b.∴a>bb-1a<2bb-1.∴a>1+1b-1a<2+2b-1.∴a>1a<4,∴a=2或a=3.∵当a=3时,由ab<a+2b得b<a,即b1<a1,与a1<b1矛盾,故a=3不合题意.∴a=3舍去,∴a=2.(Ⅱ)am=2+(m-1)b,bn=b?2n-1,...

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