bn=1 根号an 根号an 1
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证明:令cn=1/(1-an),则c1=1/(1-a1)=1,所以:c(n+1)-cn=1,是等差数列,即:cn=c1+(n-1)=n,则:an=(n-1)/nbn=[1-√a(n+1)]/n={1-
由于a1=-2,an+1=1−an1+an∴a2=1+a11−a1=−13,a3=1+a21−a2=12,a4=1+a31−a3=3,a5=1+a41−a4=−2=a1∴数列{an}以4为周期的数列∴
证明:(1)∵数列{a[n]}和{b[n]}满足a[1]=1,a[2]=2,a[n]>0,bn=√(a[n]*a[n+1]),且{b[n]}是以公比为q的等比数列∴b[1]=√(a[1]*a[2])=
(1+√2)^n第k项=Cnk*(√2)^(k-1)bn不带√2,所以k-1是偶数所以除了k=1时,后面各项都有因数2所以后面各项都是偶数k=1,Cnk*(√2)^(k-1)=11加偶数是奇数所以bn
(1)bn,√an,bn+1成等比所以an=bn*bn+1所以a1=b1*b2=3a2=b2*b3=6所以b1*(b1+d)=3(b1+d)*(b1+2d)=6解得:b1=√2d=√2/2或者b1=-
(1)由题意得,a(n-1)=an^2/4b(n-1)=lg[a(n-1)/4]=lg[an^2/16]=lg[(an/4)^2]=2lg(an/4)bn/b(n-1)=1/2为常数所以,bn是公比为
a2a3a4=15则a3=5a4=(59)÷2=7则公差d=2则a2=3,a1=1,an=2n-1bn=根号3×(1an)bn=2n×根号3b1=2根号3,b2=4根号3,b3=6根号3,则公差d=2
1=√a1a2=√2b2=b1q=√a2a3,a3=b1^2q^2/a2=q^2bn=b1q^(n-1)=√anan+1bn+2=b1q^(n+1)=√an+1an+2anan+1=2q^(n-1)a
n=√an*a(n+1)b(n+1)=√a(n+1)a(n+2)[b(n+1)/bn]^2=[a(n+1)*a(n+2)]/[a(n+1)*an]=a(n+2)/ana(n+2)=q^2*an
(1)由bn=√(4an+1)推出bn^2=4an+1即4an=bn^2-1则4a(n+1)=b(n+1)^2-1那么条件4a(n+1)=4an+2√(4an+1)+1就等价于b(n+1)^2-1=b
an,bn,an+1成等差数列,则有:2bn=an+a(n+1)由题意:a(n+1)=根号bnxb(n+1)a(n)=根号b(n-1)xb(n)将上两式代入:2bn=an+a(n+1),有2bn=根号
(1)∵数列{a[n]}和{b[n]}满足a[1]=1,a[2]=2,a[n]>0,bn=√(a[n]*a[n+1]),且{b[n]}是以公比为q的等比数列∴b[1]=√(a[1]*a[2])=√2b
1、分子有理化乘(2n+1)+√(an²+bn+1)=(4n²+4n+1-an²-bn-1)/[(2n+1)+√(an²+bn+1)]上下除n=[(4-a)n+
已知:lim[√(n^2+a*n)-(b*n+1)]=b,求a.因为√(n^2+a*n)-(b*n+1)=[√(n^2+a*n)^2-(b*n+1)^2]/[√(n^2+a*n)+(b*n+1)](分
题意:an+an+1=2bn;(1)bnbn+1=an+1*an+1(2)(2)式两边开方得:an+1=sqrt(bn)*sqrt(bn+1)(3)(1)式两边平方,展开,然后将(3)式代入,可得:b
an,bn,an+1成等差数列2bn=an+a(n+1)bn,an+1,bn+1成等比数列[a(n+1)]^2=bn*b(n+1)根据上述2式得2bn=根号(b(n-1)*bn)+根号(bnb(n+1
为了照片清晰一点,有的步骤比较简单,请见谅!
分子分母同时乘以1/an