已知an是等比数列 且公比q=2 若a1 a2 a3=a4......
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(1)由题意可知,2a3=a1+a2,即2aq2-q-1=0,∴q=1或q=-12;(II)q=1时,Sn=2n+n(n−1)2=n(n+3)2,∵n≥2,∴Sn-bn=Sn-1=(n−1)(n+2)
(1)因为a3+2是a2和a4的等差中项,所以得:2(a3+2)=a2+a4把2(a3+2)=a2+a4代入a2+a3+a4=28得:2(a3+2)+a3=282a3+4+a3=283a3=24a3=
(1)由a3=14=a1q2,以及q=-12可得a1=1.∴数列{an}的前n项和Sn=1×[1−(−12)n]1+12=2−2•(−12)n3.(2)证明:对任意k∈N+,2ak+2-(ak+ak+
我猜你的题目给出的条件是a(n+2)=a(n+1)+2an,就像楼上所列正解如下a3=a2+2a1=2a1+1a4=a3+2a2=2a1+1+2=2a1+3又an为等比数列,a2=a1*q,a3=a1
由a7=a1q6=1,得a1=q-6,从而a4=a1q3=q-3,a5=a1q4=q-2,a6=a1q5=q-1.因为a4,a5+1,a6成等差数列,所以a4+a6=2(a5+1),即q-3+q-1=
s4/s2=15/2(a4+a3+a2+a1)/(a2+a1)=15/2(a2q²+a1q²+a2+a1)/(a2+a1)=15/2[q²(a2+a1)+(a2+a1)]
因为a2+a5=9/4,a3.a4=1/2所以a2(1+q^3)=9/4,a2^2.q^3=1/2(计算过程把q^3看作整体来解)即a2=2,q=1/2所以an=4.(1/2)^(n-1)
(1)a3*a4=a2*a5=1/2a2+a5=9/4-1
一2(a3-a4)=a2-a3解得3a3=a2+2a4再除以a2得2q^2-3q+1=0解得q=1/2所以an=64*(1/2)^(n-1)
1.bn/b(n-1)=3[an-a(n-1)]=q所以an-a(n-1)=log(3)q2.a2=13a8=1d=-2an=17-2n3.n8Tn=-[a1+.an]+2[a1+.+a8=n^2-1
由a1(1+q+q^2+q^3+q^4)=242有1+q+q^2+q^3+q^4=121q(1+q)(1+q^2)=120解得q=3(一个一个试,常理这种q一定是整数.)
s5=a1+a2+a3+a4+a5=2+2q+2q*q+2q*q*q+2q*q*q*q=242q+q*q+q*q*q+q*q*q*q=120解得q=3
a2=a1*qa3=a1*q*q因为是等差数列,所以有a1+a3=2*a2a1+a1*q*q=2a1*q约去a1,得q^2-2q+1=0所以,q=1再问:q还可以等于-1/2……再答:抱歉,错了。a1
∵a1,a3,a2成等差数列∴2a1q2=a1+a1•q∴q=1或-12故选A.
(Ⅰ)依题意,得2am+2=am+1+am,∴2a1qm+1=a1qm+a1qm-1在等比数列{an}中,a1≠0,q≠0,∴2q2=q+1,解得q=1或-12.(Ⅱ)若q=1,Sm+Sm+1=ma1
2a4=-a5+a62a4=-a4q+a4q^22a4=-a4q+a4q^2a4q^2-a4q-2a4=0a4(q^2-q-2)=0a4(q-2)(q+1)=0(q-2)(q+1)=0q=2或q=-1
因为bn=根号开n次方a1a2*...an(n=1,2...)所以当n=1时,b1=a1=c,n=2,b2=根号开2次方cc*q=c根号开2次方q=cq^(0+1)/2,bn=cq(0+1+2+.+n
首先要观察到Gn是等比数列的和,其首项为a1^2,公比为q^2.等比数列的前n项和Sn=a1(1-q^n)/(1-q)则limGn/Sn=lim(a1^2(1-q^2n)(1-q))/(a1(1-q^
(1)设公比为q,2a5=4a1+(-2a3)得:q^4+q^2-2=0q不等于1,所以q=-1(2)利用分组求和法:Sn=2-2×(-1)^nAn=S1+S2+S3+……+Sn=(2+2+……+2)
a4=a1q^3a7=a4q^3……a100=a97q^3a1,a4,a7,……,a100是以a1为首项,q^3为公比的等比数列,共34项.S=a1[1-(q^3)^34]/(1-q^3)=a1(1-