1 在等比数列{an}中,已知Sn=48,S2n=60,求S3n2 已知a1=1,an+1=2(an)+3,求an
1 在等比数列{an}中,已知Sn=48,S2n=60,求S3n
2 已知a1=1,an+1=2(an)+3,求an
∵ {an}为等比数列,
∴ Sn, S2n-Sn, S3n-S2n成等比数列,即(S2n-Sn)²=Sn·(S3n-S2n)
∵ Sn=48, S2n=60,
∴ (60-48)²=48(S3n-60),解得S3n=63
1、Sn=a1 (1-q^n)/(1-q)
S2n=a1 (1-q^2n)/(1-q)
S3n=a1 (1-q^3n)/(1-q)
S2n/Sn=(1-q^2n)/(1-q^n)=(1+q^n)(1-q^n)/(1-q^n)=1+q^n=60/48=1.25
q^n=0.25
S3n/Sn=(1-q^3n)/(1-q^n)=(1-q^n)(1+q^n+q^2n)/(1-q^n)
=1+q^n+q^2n=21/16
S3n=21/16 *48=63
2、 a(n+1)=2an+3,
【a(n+1)+3】=2(an+3)
【a(n+2)+3】=2【a(n+1)+3】
【a(n+3)+3】=2【a(n+2)+3】
将(an+3)看做一个整体,bn=(an+3)
即为等比数列,公比为2
a1=1
b1=4
bn=4*2^(n-1)=2^(n+1)
an=2^(n+1)-3
1,:Sn=a1 (1-q^n)/(1-q)
S2n=a1 (1-q^2n)/(1-q)
S3n=a1 (1-q^3n)/(1-q)
S2n/Sn=(1-q^2n)/(1-q^n)=(1+q^n)(1-q^n)/(1-q^n)=1+q^n=60/48=1.25
q^n=0.25
S3n/Sn=(1-q^3n)/(1-q^n)=(1-q^n)(1+q^n+q^2n)/(1-q^n)
=1+q^n+q^2n=21/16
S3n=21/16 *48=63
2;an+1 - 2an =3
2an - 4an-1=3*2
4an-1 -8an-2 =3*4 以此类推 an+1 -2^n =3 +3*2+....3*2(N-1)次方=2(n+1)次方-2 an+1 =2(n+1)次方-2+2^n 所以an=2的n次方+2是(n-1)次方-5/2
1.在等比数列中有Sn,S2n-Sn,S3n-S2n.....成等比关系,
因为Sn=48,S2n-Sn=12,所以S3n-S2n=4,所以S3n=63.
2.因为a(n+1)=2an+3,化解得
a(n+1)+3=2(an+3)
设bn=an+3,则bn是以2为公比的等比数列,b1=a1+3=4
所以bn=4^(n-1)
所以an=bn-3=4^(n-1)-3
(1)
Sn=48=a1+a2+……+an=48
S2n=a1+a2+……an+a(n+1)+……a(2n)=60
Sn+Sn*q^n=S2n
q^n=1/4
S3n=a1+……a(3n)=S2n+a(2n+1)+……a(3n)=S2n+Sn*q^2n=60+48*(1/4)^2=63
(2)
a(n+1)=2(an)+3
a(n+1)+3=2(an)+6=2(an+3)
a1+3=4
所以{an+3}是以4为首项,2为公比的等比数列
an+3=2^(n+1)
an=2^(n+1)-3