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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...

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...

令 bn=a(n)+1 b(n+1)=a(n+1)+1 b(n+2)=a(n+2)+1 因为数列{bn}也是等比数列,所以,连续的三项: bn,b(n+1),b(n+2)也成等比, 即,首尾相乘等于中间的平方, [a(n+1)+1]^2=[a(n)+1]*[a(n+2)+1]

设等比数列{an}的公比为q,∵a1=2,∴a2=2q,a3=2q2,又{an+1}也成等比数列,∴(2q+1)2=(2+1)(2q2+1),解得q=1,∴an=2,∴Sn=2n故答案为:2n

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=...

(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...

(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...

Sn=a1(1-q^n)/(1-q) 将a1=2,q=-1/2,Sn=21/16带入得到21/16=4/3×(1-(-1/2)^n) 1-(-1/2)^n=63/64 n=6 an=2×(-1/2)^5=-1/16

∵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

说明:^——表示次方 a1=2,q>1 Sn=a1(q^n-1)/(q-1) =2(q^n-1)/(q-1) S4/S2=5 S4=5S2 2(q^4-1)/(q-1)=5×2(q^2-1)/(q-1) q^2+1=5 q^2=4 q=2 an=a1q^(n-1) a1=2 a2=2×2^(2-1)=4 a3=a2q=4×2=8 a4=a3q=8×2=16 1/a1=1/2 1/a2=1/4 1/a3=1/8 1/a4=1/16 {1...

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