已知在公比为实数的等比数列an中,a3=4,且a4,a5 4,a6,成等差数列
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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)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=
设首项为X则有X+4X+16X=21X=1通项公式an=4的(n-1)幂
/>(1)S3=a1+a2+a3=a1(1+q+q²)=a1(1+3+3²)=13a1=13/3a1=1/3an=a1q^(n-1)=(1/3)×3^(n-1)=3^(n-2)数列
a3=4a4=a3*q=4qa5=a3*q^2=4q^2a6=a3*q^3=4q^3且a4,a5+4,a6成等差数列4q+4q^3=2(4q^2+4)q^3+q-2q^2-2=0q^2(q-2)+(q
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
a9=a5+4da15=a5+10d(a5+4d)²=a5(a5+10d)8da5+16d²=10da516d²-2da5=02d(8d-a5)=0d=a5/8所以a9=
因为a5=a1+4d,a9=a1+8d,a15=a1+14d且a5a9a15成等比数列所以(a1+8d)^2=(a1+4d)(a1+14d)即(a1)^2+16a1*d+64d^2=(a1)^2+18
6m+7=3k+16(m+1)=3kk=2m+2q=bn/bn-1=an+1/an-1an+1-(an-1)=2d两个联立an-1=1+2d/q是常数所以an是常数列bn也是常数列,且bn=1
(1)S1→3=a1(1+q+q^2)=a1*(1-q^3)/(1-q)S4→6=a4(1+q+q^2)=a1*(1-q^3)/(1-q)*q^3S7→9=a7(1+q+q^2)=a1*(1-q^3)
设a(n)=a1*q^(n-1),则s(n)=a1(1-q^n)/(1-q).求出a(n-1)、s(n-1)、a(n+1)、s(n+1)并代入原不等式化简得:q^(n-2)*(1-q)0.所以q^(n
等比数列an的公比大于1,设公比为q,且q>1a1a3=6a2,a1*a2*q=6a2a1*q=6a2=6a1.a2.a3-8成等差,2a2=a1+a3-82*6=6/q+6*q-820q=6+6q^
(1).由a(m)+a(m+1)=a(k)知道3m+3(m+1)+1=3k+1,整理后有k-2m=4/3,而m,k均是N+,则k-2m也是整数,故而不存在m,k∈N+,使a(m)+a(m+1)=a(k
an=3^(n-1)S3=3b2=15b2=5b1=5-db2=5+d(a1+b1)(a3+b3)=(a2+b2)^2[(5-d)+1](9+5+d)=(3+5)^2(d+10)(d-2)=0前n项和
a5=a4*qa7=a4*q^3a6=a4*q^22(a5+a7)=a4+a62(a4*q+a4*q^3)=a4+a4*q^2a4不等于0两边同时÷a42q+2q^3=1+q^22q(1+q^2)=1
设数列{an}的公比为q(q∈R),由题意可得2(4q2+4)=4q+4q3,整理可得(q2+1)(q-2)=0,∵q∈R,∴q=2,a1=a3q2=1,∴数列{an}的通项公式为:an=2n−1,故