在等比数列an中,前n项和为sn,且sn=2的n次方-1
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![在等比数列an中,前n项和为sn,且sn=2的n次方-1](/uploads/image/f/3263189-5-9.jpg?t=%E5%9C%A8%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97an%E4%B8%AD%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E4%B8%94sn%3D2%E7%9A%84n%E6%AC%A1%E6%96%B9-1)
由Sn=3^(n+1)+r可知公比q=3取n=1得a1=9+r取n=2得a1+a2=4a1=27+r解得a1=6,r=-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+.
设{an}的公比为q,则a2=2q,a3=2q^2则(a2+1)^2=(a1+1)(a3+1)即(2q+1)^2=3(2q^2+1)解得q=1所以{an}为常数数列Sn=na1=2n
当公比为1时,Sn=n,数列{Sn+12}为数列{n+12}为公差为1的等差数列,不满足题意;当公比不为1时,Sn=1−qn1−q,∴Sn+12=1−qn1−q+12,Sn+1+12=1−qn+11−
因数列{an}为等比,则an=2qn-1,因数列{an+1}也是等比数列,则(an+1+1)2=(an+1)(an+2+1)∴an+12+2an+1=anan+2+an+an+2∴an+an+2=2a
Sn=a1(1-q^n)/(1-q)S1=a1S2=a1(1+q)S3=a1(1+q+q^2)S2+2=a1(1+q)+2S3+2=a1(1+q+q^2)+2[a1(1+q+q^2)+2]*[a1+2
新数列设为bnb1=a2=6公比变为9bn=6*9^(n-1)Sn=[6(1-9^n)]/(1-9)=[6(1-9^n)]/(-8)=[6(9^n-1)]/8=3(9^n-1)/4Sn=(9^n-1)
∵Sn=3n+a,∴a1=S1=3+a,∵an=Sn-Sn-1=(3n+a)-(3n-1+a)=2×3n-1,∴a1=2.又∵a1=S1=3+a,∴3+a=2,∴a=-1.∴an=2×3n-1.故答案
因数列{an}为等比,则an=3qn-1,因数列{an+1}也是等比数列,则(an+1+1)2=(an+1)(an+2+1)∴an+12+2an+1=anan+2+an+an+2∴an+an+2=2a
解题思路:应用特值法:Sn+1=pSn+1,分别取n=1,2,设等比数列{an}的公比为q.可得a1+a2=pa1+1,a1+a2+a3=p(a2+a1)+1,化为a1+a1q=pa1+1,p=q,又
因为6Sn=(an+1)(an+2)(1)所以6Sn-1=(an-1+1)(an-1+2)(2)(1)-(2)则an-an-1=3所以an是等差数列因为6Sn=(an+1)(an+2)可知S1=a1=
因为a(n+1)=S(n+1)-S(n)=S(n)+3n+1即a(n+1)=S(n)+3n+1(1)所以a(n)=S(n-1)+3(n-1)+1(2)(1)-(2)得a(n+1)-a(n)=S(n)-
/>首先公比q≠1,否则an=a1,矛盾,Sn=a1*(1-q^n)/(1-q)=-341a1=1,an=a1*q^(n-1)=-512即q^(n-1)=-512∴[1-(-512q)]/(1-q)=
a[n+1]=4a[n]-3n+1=4a[n]-4n+n+1因此a[n+1]-(n+1)=4a[n]-4n即b[n+1]=4b[n],也就是说b[n]是等比数列又b[1]=a[1]-1=1所以b[n]
等比数列{an},an>0,Sn=80.S(2n)=6560,Sn/S(2n)=1/(1+q^n)=80/6560q^n=81,n>1,q>1a1/(1-q)=80/(1-q^n)=-1前n项和中数值
(a2+1)²=(a1+1)(a3+1)a1=2,设an公比q(2q+1)²=3(2q²+1)4q²+4q+1=6q²+32q²-4q+2=
设公比为q,a2²=a1*a3(a2+1)²=(a1+1)(a3+1)因为a1=2所以a2²=2a3(a2+1)²=3(a3+1)解得a2=2a3=2所以sn=
已知Sn=2An-1取n=1得:S1=2A1-1又因为S1=A1,解上述方程可得:A1=1Sn=2An-1S(n-1)=2A(n-1)-1注:"n-1"为下标上下两式相减得:Sn-S(n-1)=2An
等差则2(a2+1)=(a1+1)+(a3+1)2a2=a1+a32a1q=a1+a1q²所以q²-2q+1=0q=1所以这是常数列所以an=a1=1再问:那Sn等于多少呢~~再答