已知数列an和bn满足a1=2b1=1an 1=2sn 1,数列bn为等差数列
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由bn=an-1与an-1=an[(an+1)-1]得bn=[bn+1]*(bn+1)所以bn/[bn+1]=(bn+1)所以[bn+1]/bn=1/(bn+1)即1/bn+1=(bn+1)所以{1/
(n+1)/bn=2∴bn=b1×2^(n-1)b1=a2-a1=3-1=2∴bn=2^n∴a(n+1)-an=2^n∴a2-a1=2a3-a2=2^2a4-a3=2^3……an-a(n-1)=2^(
已知:数列an满足a1=2,其前n项和为Sn=n+7-3an;数列bn满足bn=an-1,证明数列bn是等差数列.代入an=Sn-S(n-1),得Sn=n+7-3(Sn-S(n-1)),变形成:Sn-
(1)an为等比数列an=3^(n-1)Sn=n*n+2n+1n=1时b1=4n>1时bn=Sn-S(n-1)=2n+1(2)n=1时Tn=4n>1时tn=4+3^2*5+3^3*7+……+3^(n-
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
1.an-1=1/bn,an=1/bn+1a(n-1)=1/b(n-1)+11/bn+1=2-1/(1/b(n-1)+1)1/bn=1-b(n-1)/(b(n-1)+1)1/bn=1/(b(n-1)+
n=1-an,第二个式子代入bn=1-anbn+1=(1-an)/(1-an^2)=1/(1+an)an+1=1-bn+1=an/(1+an)求倒数1/(an+1)=1+1/an令cn=1/an,cn
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
an-1=an[a(n+1)-1],an[a(n+1)-2]=-1,a(n+1)=2-1/an=1+(an-1)/an,a1=2,a2=1+1/2=3/2,a3=1+(3/2-1)/(3/2)=4/3
1.a(n+1)-an=1,为定值,又a1=1,数列{an}是以1为首项,1为公差的等差数列.an=1+n-1=nn=1时,S1+b1=2b1=2b1=1n≥2时,Sn=2-bnS(n-1)=2-b(
(Ⅰ)由bn=an-1得an=bn+1代入2an=1+anan+1得2(bn+1)=1+(bn+1)(bn+1+1)整理得bnbn+1+bn+1-bn=0从而有1bn+1−1bn=1∴b1=a1-1=
an=1/(n(n+1))bn=2^n
这一看an就是等差数列,bn是等比数列,an+1-an=2,所以an=1、3、5、7、9、11、13、15、17、19……,bn=1、2、4、8、16、32、64、128……,ban的前十项和就是ba
(1)b1=√2,bn=√2*q^(n-1)(bn+1/bn)^2=an+2/an=q^2(2)Cn+1=a2n+1+2a2n+2=q*a2n-1+2q*a2n=q*(a2n-1+2a2n)=q*Cn
d(n)=2^n+n,p(1)=d(1)=2^1+1=3,p(n+1)=d(n+1)+d(n)=2^(n+1)+(n+1)+2^n+n=3*2^n+2n+1,L(2n-1)=d(2n-1)=2^(2n
1.bn=a1+a2+a3...an\nnbn=a1+a2+a3...an=n^3an=n^3-(n-1)^3=3n^2-3n+12.令a1+a2+a3...an=Snbn=b+(n-1)dbn=a1
lg(1+a1+a2+.+an)=n1+Sn=10^nSn=10^n-1n=1时,a1=S1=9n≥2时,an=Sn-S(n-1)=10^n-10^(n-1)=9*10^(n-1)n=1时,上式也成立
解an=3/4a(n-1)+1/4b(n-1)+1(1)bn=1/4a(n-1)+3/4b(n-1)+1(2)(1)+(2)得an+bn=a(n-1)+b(n-1)+2,(n>=2)),所以数列an+
a(n+1)=an+2a(n+1)-an=2an-a1=2(n-1)an=2nbn=1/[an.a(n+1)]=(1/4)(1/n-1/(n+1)]Sn=b1+b2+...+bn=(1/4)(1-1/