已知等比数列{bn}的公比为正数,数列{an}满足bn=3an

1、an=2^(n-1),b1=S1=1,bn=Sn-Sn-1=n^2-(n-1)^2=2n-1,n>=2,n=1也合适bn=2n-12、令cn=bn/an=2n-1/2^(n-1)错位相减:Sn=c

一定尽力解答,祝愉快

a1>0q>0,an=a1q^(n-1)>0,数列为正项数列.an=a1×q^(n-1)bn=log2(an)=log2[a1q^(n-1)]=log2(a1)+(n-1)log2(q)b(n+1)=

设an=a1*q^(n-1),那么bn=lgan=lg(a1*q^(n-1))=lga1+(n-1)lgq,所以b(n+1)-bn=lgq是常数,所以{bn}是等差数列

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

显然有：an=a1+(n-1)d,bn=b1*q^(n-1),又a3=b3,a7=b5,所以：a1+2d=a1*q^2,①a1+6d=a1*q^4,②由上面2个式子,得到：3①-②：2a1=a1*(3

（1）∵等比数列{bn}的公比为3∴bn+1bn=3an+13an=3an+1−an=3∴an+1-an=1∴{an}是等差数列（2）∵a1=1，an+1-an=1∴an=n则cn=1anan+1=1

(1)S5=5a1+10d=5+10d=45,d=4,a3=1+2d=9.T3=b1+b2+b3=1+q+q^2=9-q,则q=-4或q=2.因为q>0,所以q=2.{an}的通项公式为：an=1+4

/>(1)an=aⁿbn=anlgan=aⁿlg(aⁿ)=naⁿlgaSn=b1+b2+...+bn=1×a×lga+2×a²lga+...+

6m+7=3k+16(m+1)=3kk=2m+2q=bn/bn-1=an+1/an-1an+1-(an-1)=2d两个联立an-1=1+2d/q是常数所以an是常数列bn也是常数列,且bn=1

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

设an=a+n-1;bn=b*2^(n-1)a+5=4b;5(a+a+9)=15b+45a=3,b=2an=2+n,n>=1

（1）.由a（m）+a（m+1）=a（k）知道3m+3(m+1)+1=3k+1,整理后有k-2m=4/3,而m,k均是N+,则k-2m也是整数,故而不存在m,k∈N+,使a（m）+a（m+1）=a（k

（1）由题意,可得an=(1/4)^n;那么：bn+2=3*log(1/4)an=3n;所以：bn=3n-2,为等差数列；（2）由条件Cn=an*bn得到：Cn=(1/4)^n*(3n-2)=3n*(

an=3*2^(n-1)bn=lg3*2^(n-1)=lg3+lg2^(n-1)=lg3+(n-1)lg2bn-1=lg3+(n-2)lg2d=bn-bn-1=lg2

an=1000*(1/10)^(n-1)=10^3*10^(1-n)=10^(4-n)lgan=4-nbk=lga1+lga2+...+lgak=3+2+...+4-k=(3+4-k)*k/2=(7-

（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

Tn=1/a1+1/a2+……+1/anTn/q=1/a2+……+1/an+1/(q*an)Tn-Tn/q=1/a1-1/(q*an)Tn=q/a1(q-1)-1/an(q-1)

1=a1a2=r,故bn=r*q^(n-1)又b(n+1)/bn=a(n+1)*a(n+2)/(an*a(n+1))=a(n+2)/an、b(n+1)/bn=q可得当n为奇数时an=a1*q^((n+

由a/an=bn,得a/a=b,{an}是公差为d的等差数列,{bn}是公比是q的等比数列,∴a/a*an/a=q,即[a1+(n-1)d][a1+(n+1)d]/[a1+nd]^2=q(常数）对n∈