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在等比数列An中,A1=2,前n项和为Sn,若数列{An+1}也...

(1)对等比数列{bn},公比q=2+b2=1+b2.∵limn→∞Sn有意义,∴0<|q|<1,∴-4<b<0.又∵Sn=2[1?(1+b2)n]1?(1+b2),∴limn→∞Sn=21?(1+b2)=3-b.解方程21?(1+b2)=3?b,得b=4或-1.因为-4<b<0,所以b=-1.(2)当b取偶数(b=2k,k∈N*)时,{bn...

解: 设{an}公比为q,(q≠0) an=a1·qⁿ⁻¹ 设{an +1}公比为q',(q'≠0) a1+1=2+1=3 q'=[a(n+1)+1]/(an +1) =(a1·qⁿ+1)/(a1·qⁿ⁻¹+1) =q(a1·qⁿ⁻¹ +1/q)/(a1·qⁿ⁻¹+1) =q(a...

(1)设等差数列{an}的公差为d,则S5-S2=3a1+9d=27,又a1=3,则d=2,故an=2n+1;(2)由(1)可得Sn=3n+2n(n?1)2=n2+2n,由Sn,22(an+1+1),Sn+2成等比数列,∴Sn?Sn+2=8(an+1+1)2,即n(n+2)2(n+4)=8(2n+4)2,化简得n2+4n-32=0,解得n...

在等比数列{an}中,设公比q(q≠0),前n项和为Sn,当a1=1时,有4a1,2a2,a3成等差数列,∴4a2=4a1+a3,即4q=4+q2,∴q=2∴S4=1×(1?24)1?2=15.故答案为:15.

1)令A(-x,0);D(0,x) 1/2*x*(x-1)=3 x=-2(舍去),x=3 A(-3,0),D(0,3) 把ABD分别代入二次函数表达式: {9a-3b+c=0 {a+b+c=0 {c=3 解得:a=-1,b=-2,c=3 y=-x2-2x+3 2) y=-(x+1)2+4 M(-1,4) 令M'(1,4);AM'方程为y=ax+b;代入A,M' {-3a+b=0 {a+b=4 a=1,b=...

解:∵a3-a1=2 又(a3)²=a1a5 ∴(a3)²=(a3-2)a5 ∴a5=(a3)²/(a3-2) ∴1/a5=1/a3-2/(a3)² =-2[1/(a3)²-1/2a3] =-2[1/(a3)²-1/2a3+1/16-1/16] =-2(1/a3-1/4)²+1&#...

(I)∵an+1=Sn+t16…(1);an=Sn?1+t16…(2)(1)-(2)得:an+1=2an(n≥2)…(2分)∵数列{an}为等比数列,∴a2a1=2…..(4分)∵a2=S1+t16=4+t16,a1=14,∴4+t4=2,∴t=4…(6分)(II)a2=4+t16,an+1=2an(n>1),∴an+1=4+t16?2n?1(n∈N*)…....

∵2a2,S3,a4+2成等差数列,a1=1∴2S3=2a2+a4+2∴q≠1∴2×1?q31?q=2q+q3+2∴q3-2q2=0∵q≠0∴q=2∴数列{an2}是以1为首项,以4为公比的等比数列前5项和为1?451?4=341故选A

(1)∵5S1,S3,3S2成等差数列,∴2S3=5S1+3S2…(1分)即2(a1+a1q+a1q2)=5a1+3(a1+a1q),化简得 2q2-q-6=0…(2分)解得:q=2或q=-32…(3分)因为数列{an}的各项均为正数,所以q=-32不合题意…(4分)所以{an}的通项公式为:an=2n.…(5分)(2...

设等比数列的公比为q,由已知可得,a1q3-a1q2=2,4a1=3a1+a1q2,联立可解得,q=-1或q=1(舍去),a1=-1,∴an=(-1)n.Sn=?[1?(?1)n]1?(?1)=-1+(?1)n+12.

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