等差数列{An}各项均为正整数,A1=3.前n项和为Sn.等比数列{Bn}中,B1=1且B2S2=64.{Ban}是公比为64的等比...等差数列{An}各项均为正整数,A1=3.前n项和为Sn.等比数列{Bn}中,B1=1且B2S2=64.{Ban}是公比为64的等比数
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![等差数列{An}各项均为正整数,A1=3.前n项和为Sn.等比数列{Bn}中,B1=1且B2S2=64.{Ban}是公比为64的等比...等差数列{An}各项均为正整数,A1=3.前n项和为Sn.等比数列{Bn}中,B1=1且B2S2=64.{Ban}是公比为64的等比数](/uploads/image/z/2430698-50-8.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7BAn%7D%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%2CA1%3D3.%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn.%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7BBn%7D%E4%B8%AD%2CB1%3D1%E4%B8%94B2S2%3D64.%7BBan%7D%E6%98%AF%E5%85%AC%E6%AF%94%E4%B8%BA64%E7%9A%84%E7%AD%89%E6%AF%94...%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7BAn%7D%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%2CA1%3D3.%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn.%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7BBn%7D%E4%B8%AD%2CB1%3D1%E4%B8%94B2S2%3D64.%7BBan%7D%E6%98%AF%E5%85%AC%E6%AF%94%E4%B8%BA64%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0)
等差数列{An}各项均为正整数,A1=3.前n项和为Sn.等比数列{Bn}中,B1=1且B2S2=64.{Ban}是公比为64的等比...等差数列{An}各项均为正整数,A1=3.前n项和为Sn.等比数列{Bn}中,B1=1且B2S2=64.{Ban}是公比为64的等比数
等差数列{An}各项均为正整数,A1=3.前n项和为Sn.等比数列{Bn}中,B1=1且B2S2=64.{Ban}是公比为64的等比...
等差数列{An}各项均为正整数,A1=3.前n项和为Sn.等比数列{Bn}中,B1=1且B2S2=64.{Ban}是公比为64的等比数列,求An与Bn
等差数列{An}各项均为正整数,A1=3.前n项和为Sn.等比数列{Bn}中,B1=1且B2S2=64.{Ban}是公比为64的等比...等差数列{An}各项均为正整数,A1=3.前n项和为Sn.等比数列{Bn}中,B1=1且B2S2=64.{Ban}是公比为64的等比数
等差数列{an}各项均为正整数,a1=3;设公差为d;
an=a1+(n-1)d=3+(n-1)d;
前n项和为Sn=3n+n(n-1)d/2;
等比数列{bn}中,b1=1;设公比为q;
bn=b1q^(n-1)=q^(n-1);
b2S2=64;
q*(6+d)=64;
ban=q^(an-1)=q^[2+(n-1)d];
ba(n-1)=q^[a(n-1)-1]=q^[2+(n-2)d]
{ban}是公比为64的等比数列;
ban/ba(n-1)=q^d=64;
因为an,bn各项都是正整数,所以d,q为正整数;
64=2^6=4^3=8^2=64^1;
q,d为(2,6)(4,3)(8,2)(64,1)
只有(8,2)满足q*(6+d)=64;
所以d=2;q=8;
an=3+2(n-1);
bn=8^(n-1)=2^3(n-1);
an=a1+(n-1)d=3+(n-1)d;
前n项和为Sn=3n+n(n-1)d/2;
等比数列{bn}中,b1=1;设公比为q;
bn=b1q^(n-1)=q^(n-1);
b2S2=64;
q*(6+d)=64;
ban=q^(an-1)=q^[2+(n-1)d];
ba(n-1)=q^[a(n-1)-1]=q^[2+(n-2)d]
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an=a1+(n-1)d=3+(n-1)d;
前n项和为Sn=3n+n(n-1)d/2;
等比数列{bn}中,b1=1;设公比为q;
bn=b1q^(n-1)=q^(n-1);
b2S2=64;
q*(6+d)=64;
ban=q^(an-1)=q^[2+(n-1)d];
ba(n-1)=q^[a(n-1)-1]=q^[2+(n-2)d]
{ban}是公比为64的等比数列;
ban/ba(n-1)=q^d=64;
因为an,bn各项都是正整数,所以d,q为正整数;
64=2^6=4^3=8^2=64^1;
q,d为(2,6)(4,3)(8,2)(64,1)
只有(8,2)满足q*(6+d)=64;
所以d=2;q=8;
an=3+2(n-1);
bn=8^(n-1)=2^3(n-1);
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