设等差数列an的首项为1,其前n项和sn ,bn是公比为正整数的等比数列,其首项为3 ,前n项和为tn ,若a3+b3=17,t3-s3=12.求an和bn的通项公式!
人气:297 ℃ 时间:2019-10-24 12:48:31
解答
设{an}公差为d,{bn}公比为q,q>0a3+b3=a1+2d+b1q^2=1+2d+3q^2=17d=(16-3q^2)/2T3-S3=b1(1+q+q^2)-(a1+a1+d+a1+2d)=3q^2+3q+3-3-3d=123q^2+3q-3d=12d=(3q^2+3q-12)/3=q^2+q-4(16-3q^2)/2=q^2+q-4整理,得5q^2+2q-24=0(5...太给力了,你的回答已经完美的解决了我问题!
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