已知正项数列{an},{bn}满足：对任何正整数n,都有an,bn,a(n+1）成等差数列,bn,a(n+1),b(n+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}通用公式
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：323 ℃ 时间：2020-05-24 15:54:21
解答
1.令Cn=根号Bn 则由bn,a(n+1),b(n+1)成等比数列知,AN+1=CN乘以CN+1由an,bn,a(n+1）成等差数列,2CN平方=AN + AN+1 = CN-1乘以CN + CN乘以CN+1即2CN=CN-1 + CN+1 亦即 2根号BN=根号BN-1 + 根号BN+1证毕2.根据第一问知,...AN=根号BN 乘以 根号bn+1 =cn 乘以 cn+1 =（N+4)(N+5)除以2那么当N=1的时候，A1=15但题目说A1=10与事实不符。不小心弄错了~An=(n+3)(n+4)\2Bn=(n+4)的平方/2
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