等差数列AN各项均为正整数
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(1)设{an}公差为d,{bn}公比为q,则an=3+(n-1)d=dn+3-d,bn=q^(n-1),所以b(an)=q^(an-1)=q^(dn+2-d),因为{b(an)}的公比为64,所以q
a(n+1)=√[bn*b(n+1)]2bn=an+an+12bn=√[bn*b(n-1)]+√[bn*b(n+1)]2√bn=√b(n-1)+√b(n+1)所以数列{√bn}为等差数列√b1=√2(
设等差数列的公差为d,则a3=a5-2d=6-2d,an1=a5+(n1-5)d=6+(n1-5)d.∵a3,a5,an1成等比数列,∴a52=a3an1化简即(6n1-42)d-2(n1-5)d2=
d=50/(n-1),注意,题目中说个各项均为正整数,所以d也只能是正整数,因此,d只能是1,2,5,10,25,50这些数,此时n分别为51,26,11,6,3,2n+d最小就是16
由基本不等式得:a3+a9≥2√(a3*a9)=2*a6=2*b7又因为b7为b4和b10的等差中项,所以2*b7=b4+b10所以a3+a9≥b4+b10当a3=a9时取等号
1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=
sn=an(an+1)/2s(n-1)=a(n-1)(a(n-1)+1)/2两式相减an=an(an+1)/2-a(n-1)(a(n-1)+1)/2an^2-an-a^2(n-1)-a(n-1)=0(
可用递推法:2Sn=An+An*An递推2Sn-1=An-1+An-1*An-1两市相减,得:An+An-1=An*An-An-1*An-1因为An为正数,所以An-An-1=1之后求An,然后用求和
sn=(1/8)(an+2)²S(n-1)=(1/8)[a(n-1)+2]²an=Sn-S(n-1)=(1/8){(an+2)²-[a(n-1)+2]²}=(1
(1)设{an}的公差为d,{bn}的公比为q,则d为正整数,an=3+(n-1)d,bn=qn-1依题意有ban+1ban=q2+ndq2+(n-1)d=qd=64,且S2b2=(6+d)q=64,
2a3+2a11=2(a3+a11)=4a7,因此4a7-a7^2=0,所以a7=4,所以b6*b8=b7^2=a7^2=16,选D根据题意可得4S2=S1+3S3,即4(a1+a1q)=
a1+a10=a2+a9=.a3+a8=10>=2更号下A3*A8A3*A8=25
an=a1+(n-1)d=1949+(n-1)d=2009(n-1)d=2009-1949=60(n-1)+d≥2√[d(n-1)]当n-1=d时,取最小值所以,已知(n-1)d=60,求n+d=(n
(1)an,bn^2,an+1成等差数列2bn^2=an+a(n+1)bn^2,an+1,bn+1^2成等比数列a(n+1)^2=bn^2*b(n+1)^2a(n+1)=bnb(n+1)2bn^2=a
由题意设等比数列{an}的公比为q(q>0),∵a1,12a3,a2成等差数列,∴2×12a3=a1+a2,∵a1≠0,∴q2-q-1=0,解得q=1+52或q=1−52(舍去).∴a3+a4a4+a
an,bn,an+1成等差数列2bn=an+a(n+1)bn,an+1,bn+1成等比数列[a(n+1)]^2=bn*b(n+1)根据上述2式得2bn=根号(b(n-1)*bn)+根号(bnb(n+1
由a1=1,得到an=a1+(n-1)d=1+(n-1)d=51,即(n-1)d=50,解得:d=50n−1,因为等差数列的各项均为正整数,所以公差d也为正整数,因此d只能是1,2,5,10,25,5
k=b1+(k-1)d(d为公差,常数)设An=a1*q^(n-1)(q为公比,常数)则Abk=a1*q^[b1+(k-1)d]Ab(k-1)=a1*q^[b1+(k-2)d]所以Abk:Ab(k-1
Sn、an、1成等差,则2an=Sn+1(n=1时,得a1=1),当n≥2时,有2a(n-1)=S(n-1)+1,则2an-2a(n-1)=an,即an/[a(n-1)]=2=常数,所以{an}是等比
由题意知2an=Sn+1/2,an>0,当n=1时,2a1=a1+1/2,解得a1=1/2,当n≥2时,Sn=2an-1/2,S(n-1)=2a(n-1)-1/2,两式相减得an=Sn-S(n-1)=