等比数列an满足:a1 a6=11,a3*a4=32 9
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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=
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^
S1=a1(1-q)/(1-q),S2=a1(1-q^2)/(1-q),...,Sn=a1(1-q^n)/(1-q).S1+S2+...+Sn=[a1/(1-q)]*[1-q+1-q^2+...+1-
(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)=√[bn*b(n+1)]2bn=an+an+12bn=√[bn*b(n-1)]+√[bn*b(n+1)]2√bn=√b(n-1)+√b(n+1)所以数列{√bn}为等差数列√b1=√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
Sn=a1(1-qn)/(1-q)[q不等于1];q=1时,Sn=n*a11)q=1,lim(Sn-a1)=lim[a1(n-1)]与题意矛盾;2)q>1,左边lim(Sn-a1)=-a1=1/2,即
lim(a1+a2+a3+...+an)=1/2说明等比数列为收敛数列,即公比q0Sn=a1(1-q^n)/(1-q)limSn=a1/(1-q)=1/2a1=1/2-1/2q因为0
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
设an=1+d(n-1)a1*a4=a2*a2故1*(1+3d)=(1+d)(1+d)解上面的方程得d=0或1(0舍去)故d=1an=n
∵a2*a4=4∴a3=2.q=1/2.an=2^(4-n)2^(9-3n)>1/9.9-3n>=-3n
a1(q+q^3)=4a1(1+q+q^2)=14两式相除:(q+q^3)/(1+q+q^2)=2/7求得qan+an+1+an+2=(a1+a2+a3)*q^(n-1)>1/9关键是求q说实在的,我
当n=1时,a1a2=16①;当n=2时,a2a3=256②,②÷①得:a3a1=16,即q2=16,解得:q=4或q=-4,当q=-4时,由①得:a12×(-4)=16,即a12=-4,无解,所以q
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
设[an+1+p(n+1)+q]/[an+pn+q]=m得an+1+p(n+1)+q=man+mpn+mq.又an+1=2an+n+1,则2an+n+1+pn+p+q=man+mpn+mq,即(2-m
an-a[n-1]=1/2^na[n-1]-a[n-2]=1/2^(n-1)...a2-a1=1/2^2以上各式相加得:an-a1=(1/2^2+...+1/2^n)=1/2^2*(1-1/2^(n-
(1)bn+1=(an+1-2)/(1-an+1)=(an-2)/(2-2an)bn=(an-2)/(1-an)bn+1/bn=1/2b1=-1/2bn为等比数列(2)(an-2)/(1-an)=-1
(1)证明:由条件得a[n+2]-a[n+1]=2(a[n+1]-a[n])首项为a[2]-a[1]=5-2=3,公比为2,所以{a[n+1]-a[n]}为等比数列由(1)得a[n+1]-a[n]=3