一直等比数列an 首项为81数列bn满足bn=log3an
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A(n+1)=2S(n)+1,A(n)=2S(n-1)+1,A(n+1)-A(n)=2[S(n)-S(n-1)]=2[A(n)],A(n+1)=3A(n)所以,数列{A(n)}是首项为1,公比为3的等
A1=1A2-A1=A1*1/3=1/3..An-A(n-1)=A1*(1/3)^(n-1)=1/3^(n-1)左右两边分别相加:左边=A1+A2-A1+..+An-A(n-1)=An=1+1/3+.
由题意可知,Sn=1-q∧n/1-q.Sn-1=1-q∧n-1/1-q.an=Sn-Sn-1=q∧n-1.所以1/an=1/q∧n-1.所以Sn=1+1/q+1/q²+1/q³+.
n=1/anan=q的n-1次方bn=q的1-n次方bn=1+1/q+1/q²+…1/q的n-1次方bn的前n项和=(1-(1/q)的n次)/(1-1/q)
缺少条件,{an}为正项数列,否则log3(an)无意义,题目没法解.证:数列为正项数列,公比q>0a(n+1)/an=qb(n+1)-bn=log3[a(n+1)]-log3(an)=log3[a(
(1)等比数列{an},首项为81设an=a1*q^(n-1)=81*q^(n-1)数列{bn}满足bn=log3为底an∴bn=log3为底[81*q^(n-1)]=log3为底81+log3为底q
1设an工笔qbn-bn-1=log3an-log3an-1=log3(an/an-1)=log3q=d所以bn为等差数列2b1=log3a1=4由题意可知d
(1)是首项Ak+1公比q(2)是首项A2公比q^2(3)是首项A1公比q^10
是等比数列.奇数项a1,a3,a5,.,公比为q².每隔10项取出一项也等比,a1,a11,a21,...,公比为q^10一般地,每隔m项取出一项成等比(m∈N*),即a1,a(m+1),a
证:设等比数列{an}公比为q,对于数列{bn},对数有意义,q>0an=a1×q^(n-1)n=1时,b1=log3(a1)=log3(81)=4n≥2时,bn=log3(an)=log3(a1×q
Sn=4An-3S(n-1)=4A(n-1)-3Sn-S(n-1)=An=4An-3-[4A(n-1)-3]=4an-3-4A(n-1)+3=4An-4A(n-1)3An=4A(n-1)An/A(n-
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
sn=a1*(1-q^n)/(1-q)带入a1=1,q=a-3/2,sn=a(n无穷大)(1-(a-3/2)^n)/(5/2-a)=a因为(a-3/2)^n当n无穷大时存在,所以有-1
已知等差数列{an}的首项为a1,公差为d,前n项和为Sn,an=a1+(n-1)d,Sn=n(a1+an)/2①数列{(1/2)^an}为等比数列;(1/2)^an/(1/2)^a(n-1)=(1/
数列{an}前N项和Sn3Sn=(an-1),(1)当n>=2,有:3Sn-1=[a(n-1)-1],(2)(1)-(2),3an=an-an-1an/an-1=-1/2,(n>=2)当n=1,3S1
s1=as2=2a+ds3=3a+3ds2^2=s1*s3(2a+d)^2=a*(3a+3d)4a^2+4ad+d^2=3a^2+3ada^2+ad+d^2=0再解方程求出d.
an+Sn=2n令n=1a1+S1=2=>a1=1又a(n-1)+S(n-1)=2(n-1)与上式作差an-a(n-1)+an=22an-a(n-1)=2an-2=(1/2)[a(n-1)-2]得证a
∵Sn=kq^n-k∴S(n+1)=kq^(n+1)-k∴a(n+1)=S(n+1)-Sn=[kq^(n+1)-k]-(kq^n-k)=k[q^(n+1)-q^n]=k[(q-1)q^na(n+1)/
a2=a1+da4=a1+3da2^2=a1a4a2^2=(a1+d)^2=a1^2+2a1d+d^2a1a4=a1(a1+3d)=a1^2+3a1da1^2+2a1d+d^2=a1^2+3a1da1
设等差数列{an}的公差为d,等比数列{bn}的公比为q,由题意d为正整数,又a1=3,b1=1,所以an=3+(n-1)d,bn=qn-1--------(6分)又因为数列{ban}是公比为64的等