Question

求解一道高一数学题..(数列)..追加分

(1)已知数列{an}的前n项和为Sn=b×2^n+a(a≠0,b≠0)若数列{an}是等比数列,则a b 应满足的条件为
(A)a-b=0 (B)a-b≠0 (C)a+b=0 (D)a+b≠0

(2)在等比数列{an}中,a1-a5=-15/2,S4=-5,则a4=?

(3)已知数列{an}满足a1=1,a2=r(r>0),数列{bn}是公比为q的等比数列(q>0),bn=an+a(n+1),cn=a(2n-1)+a2n,求cn

Answer 1

Sn=b×2^n+a
等比数列的Sn=[a1(1-q^n)]/(1-q)
=a1/(1-q)-a1q^n/(1-q)
对照一下
a=a1/(1-q)
b=-a1/(1-q)
那么a+b=0 b不是0

a1-a5=-15/2,S4=-5
a1-a1q^4=-15/2=a1(1-q^4)
a1(1-q^4)/(1-q)=-5
[-15/2]/(1-q)=-5
q=-0.5
a1=-4

a4=a1q^3=0.5

(3)已知数列{an}满足a1=1,a2=r(r>0),数列{bn}是公比为q的等比数列(q>0),bn=an+a(n+1),cn=a(2n-1)+a2n,求cn
b1=1+r
bn=(1+r)q^(n-1)=an+a(n+1)
a1+a2=(1+r)q^0
a3+a4=(1+r)q^2
...

a(2n-1)+a2n=(1+r)q^(2n-2)=Cn

Answer 2

第一题我感觉选B，因为q不一定为2