已知等比数列an满足 a1 a2 a3 a4 a5=3
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用数学归纳法a1=1/2a2=3a1+1=5/2a3=3a2+1=17/2a1+1/2=1a2+1/2=3a3+1/2=9因此先猜想a[n+1]+1/2=3(an+1/2)已证n=2,3时成立假设n=
An=A1+(A2-A1)++(An-An-1)=1(1-(1/3)^n)/(1-1/3)=2/3-2(1/3)^(n+1)
a2a4=a3*a3=144a2+a4=30a2=6a4=24q=2a1=3an=3*2^(n-1)或者a2=24a4=6q=1/2a1=48an=48*(1/2)^(n-1)数列{an}单调递增q>
1、证:a(n+1)=3an+2a(n+1)+1=3an+3[a(n+1)+1]/(an+1)=3,为定值.a1+1=1+1=2数列{an+1}是以2为首项,3为公比的等比数列.2.an+1=2×3^
a3^2+2a3*a5+a5^2=49(a3+a5)^2=49a3+a5=7再问:-7把再答:嗯忘看了an
lgan=3n+5an=10^(3n+5)a(n+1)=10^(3n+8)a(n+1)/an=10^3所以an是等比数列
(1)∵a(n+1)=2an+1∴a[n+1]+1=2a[n]+2=2(a[n]+1)∴a[n]+1为等比数列,等比=2(2)a[n]+1=(a[1]+1)*2^(n-1)=2^n∴a[n]=-1+2
a(n+1)=2an-n+1a(n+1)=2an-2n+(n+1)a(n+1)-(n+1)=2(an-n)∴{an-n}是公比为2,首项为2-1=1的等比数列an-n=1×2^(n-1)=2^(n-1
a(n+1)+1/2=3an+1+1/2=3(an+1/2)a1+1/2=1所以{an+1/2}是以1为首相,3为公比的等比数列an+1/2=3^(n-1)an=3^(n-1)-1/2
An=3乘以2的n-1次方
令Sn为an前n项和,Sn=n-an,S(n-1)=n-1-a(n-1),两式相减,an=1-an+a(n-1),2(an-1)=a(n-1)-1,所以an-1是公比为1/2的等比数列,a1-1=-1
lgAn-lgA(n-1)=lg[An/A(n-1)]=3n+5-3(n-1)-5=3所以An/A(n-1)=1000所以是等比数列再问:谢了袄哥们再答:不谢,要互相帮助
a(n+1)/an=10∧[(3n+8)-(3n+5)]=10∧3再问:那为什么a(n-1)=10^(3n+2)回答这个之后马上好评求解!!再问:或者a(n+1)=10^(3n+8)再问:懂了!!
∵an=a1•qn-1∴13=98•(23)n−1∴n=4故答案是4
Sn=2n-an,(1)S(n+1)=2*(n+1)-a(n+1)(2)(2)-(1)得:a(n+1)=2-a(n+1)+an.即:2*a(n+1)=2+an.变形:2*[a(n+1)-2]=an-2
a(n+1)+1=2an+2=2(an+1)[a(n+1)+1]/(an+1)=2所以an+1是等比数列[a(n+1)+1]/(an+1)=2则q=2所以an+1=(a1+1)*2^(n-1)=2^n
n=b1.q^(n-1)bn=an-3nan=bn+3n=b1.q^(n-1)+3nSn=a1+a2+...+an=b1(q^n-1)/(q-1)+3n(n+1)/2
设公比为q,…(1分)由已知得 a1+a1q2=10a1q3+a1q5=54…(3分)②即a1(1+q2)=10a1q3(1+q2)=54…(5分)②÷①得 q3=18,即q=12
设a2=a,a3=aq,a4=aq^2,a5=aq^3,a6=aq^4a2*a4+2a3*a5+a4*a6=a*aq^2+2aq*aq^3+aq^2*aq^4=a^2(q^2+2q^4+q^6)=a^
已知等比数列an,首项为81,数列bn满足bn=log3an,其前n项和sn(1)证明:bn-b(n-1)=log(3)an-log(3)an-1=log(3)an/a(n-1)=log(3)q∵b1