已知an是递增的等差数列a2 *a4=3,a1 a5=4
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an=3n,bn=2^(n-1)分式上下同时乘以2,把2bn化成b(n+1),另s=b(n+1),则cn=s/[(s+1)(s+2)]=s/(s+1)-s/(s+2),另dn=bn/(bn+1),则c
(1)由题意可知,2a3=a1+a2,即2aq2-q-1=0,∴q=1或q=-12;(II)q=1时,Sn=2n+n(n−1)2=n(n+3)2,∵n≥2,∴Sn-bn=Sn-1=(n−1)(n+2)
a1,a2,a3,a4,a5,a6,a7中取走任意三项的方法=7C3=35因为an是单调递增的等差数列,因此唯有次序的跳跃选取,或不跳跃的选取,才能是等差数列所以不跳跃式的选取3个(如a3,a4,a5
n=(a1+2a2+...+nan)/(1+2+...+n)a1+2a2+...+nan=(1+2+...+n)bn=n(n+1)bn/2(1)a1+2a2+...(n-1)an=n(n-1)b(n-
∵an是等差数列∴2a2=a1+a3∵a1+a2+a3=153a2=15a2=5∴a1+a3=10a3=10-a1∵a1+1.a2+3.a3+9成等比数列∴(a2+3)^2=(a1+1)(a3+9)6
a3/a1=a4/a3即为:(a1+2d)/a1=(a1+3d)/(a1+2d)因为d=2,即为a1²+,8a1+16=a1²+6a1即得a1=-8故a2=-8+2=-6
(1)a2*a4=(a1+d)(a1+3d)=3a1+(a1+4d)=4解得,d=1a1=0∴an=n-1Sn=n(n-1)/2(2)∵b1/3+b2/3^2+.+bn/3^n=a(n+1)∴b1/3
A2+A5=A3+A4=24,A2*A5=108A2=6A5=18AN=4N-2再问:非常感谢,可以继续帮我答一下吗?再答:(2)TN+1/2BN=TN+1/2(TN-Tn-1)=3/2*Tn-1/2
(1)a3+a4=24等价于2a1+5d=24.a2*a5=108等价于a1^2+5a1d+4d^2=108.解出a1和d.楼主亲自算一下吧,培养计算能力.(2)Tn=1-(1/2)bn……[1]Tn
a1*p=a2a1*p^3=a4,a1*p-a1=a1*p^3-a1*Pp-1=p^(p^2-1);(p-1)(p*(p+1)-1)=0,p=1,或p^2+p-1=0,p=(-1+√5)/2,p=(-
a1,a2,a4成等差数列2a2=a1+a4即2a1*q=a1+a1q^3a1不为0所以:2q=1+q^3q^3-2q+1=0q^3-q^2+q^2-2q+1=0q^2*(q-1)+(q-1)^2=0
a1,a2,a4成等差数列所以2a2=a1+a4{an}是等比数列a2=a1qa4=a1q^3所以2×a1q=a1+a1q^3即:q^3-2q+1=0(q-1)(q^2+q-1)=0q=1或q=(-1
a2=a1qa8=a1q^7a5=a1q^42a8=a2+a52a1q^7=a1q+a1q^42q^6=1+q^32q^6=1+q^32q^6-q^3-1=0(2q^3+1)(q^3-1)=0q^3=
这个是今年广东的高考试题,应该是题目出错了正确的题目是:已知递增的等差数列{an}满足a1=1,a3=a2^2-4,则an=____.答案:公差d=2an=2n-1
A3+A1=2A2A2的平方-3=2A2A2的平方-2A2-3=0∵A2>0∴A=3∴d=2∴An=1+2(n-1)=2n-1再答:A3+A1=2A2A2的平方-3=2A2A2的平方-2A2-3=0∵
再答:亲我看不懂Cn是多少再答:能说清楚问题吗再答:已通知提问者对您的回答进行评价,请稍等再问:Cn等于anbn再答:再答:不懂再问再问:这样就够了,谢谢
(1)设公比为q由题意得:a2=4,∵2(a3-3)=a2-1+a4-9,∴2(4q-3)=3+4q2-9,解得:q=2∴an=2n(2)∵Sn=b1+b2+…+bn=1×2+2×22+…+n×2n∴
设等差数列{an}的公差为d,(d>0)则1+2d=(1+d)2-4,即d2=4,解得d=2,或d=-2(舍去)故可得an=1+2(n-1)=2n-1,Sn=n(1+2n−1)2=n2,故答案为:2n
设公比为q,数列是单调递增等比数列,则首项a1>0,公比q>1a3+2是a2、a4的等差中项,则2(a3+2)=a2+a4a2+a3+a4=2(a3+2)+a3=3a3+4=283a3=24a3=8a