河南省焦作市2023-2024学年(下)八年级期中诊断试卷数学.考卷答案

河南省焦作市2023-2024学年(下)八年级期中诊断试卷数学.考卷答案试卷答案,我们目前收集并整理关于河南省焦作市2023-2024学年(下)八年级期中诊断试卷数学.考卷答案得系列试题及其答案,更多试题答案请关注本网站↓↓↓

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河南省焦作市2023-2024学年(下)八年级期中诊断试卷数学.考卷答案试卷答案

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分析(I)由于S6=9S3,可得q≠1,于是$\frac{{q}^{6}-1}{q-1}$=$\frac{9({q}^{3}-1)}{q-1}$,化简解得q即可得出.
(II)a1b1+a2b2+…+anbn=(n-1)×2n+1,可得当n=1时,b1=1;利用递推关系即可得出anbn=n•2n-1,即可得出.

解答解:(I)∵S6=9S3,∴q≠1,
∴$\frac{{q}^{6}-1}{q-1}$=$\frac{9({q}^{3}-1)}{q-1}$,
化为q3+1=9,解得q=2.
∴an=2n-1
(II)∵a1b1+a2b2+…+anbn=(n-1)×2n+1,
∴当n=1时,b1=1;
当n≥2时,a1b1+a2b2+…+an-1bn-1=(n-2)×2n-1+1,
∴anbn=n•2n-1,又an=2n-1
∴bn=n,n=1时也成立.
∴bn=n.
∴数列{bn}的前n项和Tn=$\frac{n(n+1)}{2}$.

点评本题考查了递推公式、等差数列与等比数列的通项公式及其前n项和公式,考查了推理能力与计算能力,属于中档题.