A1 A2 A3 ..... An=B怎么套函数?

等比数列A2=A1qA3=A1q^2A1+A2+A3=7所以A1+A1q+A1q^2=7AA1A2A3=8A1^3q^3=8A1q=2代入A得BA1+2+2q=7A1=5-2q代入(5-2q)q=2-

设公比为q.a1+a2=a1(1+q)=9a1a2a3=a1^3q^^3=27a1q=3a1=3/q(3/q)(1+q)=93q+3=9q6q=3q=1/2a1=3/(1/2)=6Sn=a1(1-q^

a1a2a3=5⇒a23=5；a7a8a9=10⇒a83=10，a52=a2a8⇒a65＝a32a38＝50⇒a4a5a6＝a35＝52，故选A．

8设第一项a.等比为q则a（1＋q+q^2)=-3a^3q^3=8a＝-1q=-2A4=aq^3=-1*-2^3=8

由等比数列性质得到a1a3=a2的平方a1a2a3=a2的立方=8a2=2a1+a2+a3=a2/q+a2+a2*q=-3q=-2q=-1（不满足舍去）当q=-2时候a4=a2*q的平方=2*4=8

因为a1a2a3=8所以a2/q*a2*a2*q=8a2^3=8,a2=2又a1+a2+a3=7即a2/q+a2+a2*q=71/q+q=5/2=2+1/2所以q=2或1/2即a1=1或4.所以an=

因为a1+a2+a3=7,a1a2a3=8又因为等比数列{an},那么a2*a2=a1a3,那么a1a2a3=a2a2a2=8,所以a2=2,那么a1+a3=5,同时a1a3=4所以a1=1,a3=4

a1=a1a2=a1qa3=a1q^2a1(1+q+q^2)=14a1a2a3=a1^3q^3=64a1q=4a1=4/q代入,4(1+q+q^2)=14q整理,得2q^2-5q+2=0(q-2)(2

a1a2a3=1a2^2=a1*a3a2^3=1a2=1a4=4a2+a4+a6+...+a2n是以a2=1为首项公比q=4的等比数列项数为na2n=a2*4^(n-1)=4^(n-1)再问：这是选择

才2个条件是求不出的,需加多一个条件.如为等差数列,得3a2=-3,a2(a2-d)(a2+d)=8,得a2=-1,d=3或-3,{a1,a2,a3}={-4,-1,2}如为等比数列,得a2/q+a2

因为a1+a2+a3=7,a1a2a3=8又因为等比数列{an},那么a2*a2=a1a3,那么a1a2a3=a2a2a2=8,所以a2=2,那么a1+a3=5,同时a1a3=4所以a1=1,a3=4

1

∵a1a3=a2的平方,第二式得a2=6一式为a2/q+a2+a2q=1,得6q²+5q+6=0∴△=5²-4x6x6=-119＜0无解

a1+a1q+a1q^2=7a1^3q^3=8a1q=2a1+2+a1q^2=7a1+a1q^2=5a1=2/q2/q+2/q*q^2=52/q+2q=52+2q^2=5q2q^2-5q+2=0(2q

因为a1+a2+a3=7,a1a2a3=8又因为等比数列{an},那么a2*a2=a1a3,那么a1a2a3=a2a2a2=8,所以a2=2,那么a1+a3=5,同时a1a3=4所以a1=1,a3=4

设b1=a1a4a7...a28;b2=a2a5a8...a29;b3=a3a6a9...a30,则有b3=b2*2^10=b1*2^20,所以b2=2^(30/3)=2^10,故b3=2^20,即答

⑴若a1＋a2＋a3=21,a1a2a3=216,设a1=a2/q,a3=a2qa2/q＋a2＋a2q=21a2³=216=6³a2=66/q＋6＋6q=211/q＋q=5/2=1

题为：在数列{a[n]}中,a[1]=2,a[n+1]=a[n]+cn(c是常数),且a[1]、a[2]、a[3]成等比数列,求数列{(a[n]-c)/(n.c^n)}的前n项之和T[n].其中[&n

迅速算的话,可以这样算：a1a2a3=a2(a1a3)=a2^3=27,a2=3a1+a2+a3=a2(1+q+1/q)=13,即q+1/q=10/3,得q=3所以an=3^(n-1)

a1a2a3…an=n^2n=1时a1=1n>1时a1a2a3…an=n^2(1)a1a2a3…a(n-1)=(n-1)^2(2)(1)/(2)=an=n^2/(n-1)^2a3=9/4a5=25/1