已知正项数列{an}{bn}满足,对任意正整数n,都有an,bn,an+1成等差数列,bn,an+1,bn+1成等比数列
且a1=10,a2=15
求证:数列(根号Bn)是等差数列
求数列{an},{bn}通项公式
设Sn=1/(a1)+1/(a2)+1/(a3)+.1/(an)如果对任何正整数n,不等式2aSn
人气:423 ℃ 时间:2020-05-22 01:09:22
解答
1.证明:因为bn,a(n+1),b(n+1)成等比数列,所以[a(n+1)]²=bnxb(n+1)(n∈N*)a(n+1)=√[bnxb(n+1)] 所以an=√[bnxb(n-1)] (n≥2)因为an,bn,a(n+1)成等差数列,所以2bn=an+a(n+1) (n∈N*)所以2bn=√[bnxb(...
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