www.1862.net > 在等比数列{An}中,(1)A4=2,A7=8,求An,(2...

在等比数列{An}中,(1)A4=2,A7=8,求An,(2...

(1)q^3=a7/a4=8/2=4 q=4^(1/3)=2^(2/3) a1=a4/q^3=2/4=1/2 an=a1*q^(n-1)=1/2*(2^(2/3))^(n-1)=2^(-1+2n/3-2/3)=2^(2n/3-5/3) (2) a2+a5=a2(1+q^3)=18 a3+a6=a3(1+q^3)=9 下式/上式得:q=a3/a2=1/2 a2+a5=a1*1/2+a1*(1/2)^4=18 a1=32 an=a1*q...

(1)数列{an}是等比数列,由于a4+a7=2,a5a6=-8根据等比数列的性质:a5a6=a4a7=-8所以:a4+a7=2a4a7=?8,解得:a4=4a7=?2或a4=?2a7=4①当a4=4a7=?2时,利用an=a4qn?4,解得:an=4?(?12)n?43②当a4=?2a7=4时,利用an=a4qn?4,解得:a...

1,a4/a2=4,所以公比是2,a1=1an=a1x2的n-1次方

q=a4/a3=2/5 a7=a1*q^6=(2/5)^6

(Ⅰ) 设等比数列{an}的公比为q,∵a4=a1q3,∴q=2.∴an=n?1.…(6分)(Ⅱ)由题意得abn2=a2?an+2,∴(2bn?1)2=2?22n,得bn=n+2,∵bn+1-bn=1,∴数列{bn}是以首项为3,公差为1的等差数列.…(9分)∴b1+b2+b3+…+bm≤=(m+5)m2≤b10=12,…(11分)即...

解: a4=a1q³ q³=a4/a1=-4/½=-8 q=-2 |a1|=|½|=½ |a(n+1)/an|=|q|=|-2|=2,为定值 数列{|an|}是以½为首项,2为公比的等比数列。 |a1|+|a2|+...+|an| =½(2ⁿ-1)/(2-1) =(2ⁿ-1)/2

由等比数列得性质可得a1a10=a2a9=…=a5a6,又∵a5a6+a4a7=8,∴a1a10=a2a9=…=a5a6=4,∴log2a1+log2a2+…+log2a10=log2(a1?a2?…a10)=log2(a1a10)5=log245=log2210=10故答案为:10

解:a3=1,a4=5/2,得 q³=5/2 a7=a4*q³=25/4

(1)设公比为q,由a4,a5+1,a6成等差数列,得2(a5+1)=a4+a6,即2(1q2+1)=1q3+1q,整理得(q2+1)(2q-1)=0,解得q=12,所以an=a7qn?7=12n?7;(2)由(1)知,a1=26=64,所以Sn=64(1?12n)1?12=128(1?12n);

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