等差数列<an>的各项为正数,a1=3,前N项和为sn,<Bn>为等比数列,B1=1,且b2S2=64,b3S3=960~问〔1〕求an与bn注意要详解…
人气:148 ℃ 时间:2019-11-04 13:37:37
解答
因为an等差,所以设等差数列公差为d,等比数列公比为q,则有 a2=3+d,a3=3+2d 由b2s2=64和b3s3=960知道 q(3+a2)=64,q^2 *(3+a2+a3)=960 解得:d=2,q=8故:an=3+(n-1)*2=2n-1bn=8^(n-1)Sn=3n+n(n-1)=n^2+2n=n(n+2)...
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