Question

已知等比数列an满足a2-a3的绝对值等于10

Answer 1

已知等比数列{an}满足:|a2-a3|=10,a1a2a3=125
(1)求数列{an}的通项公式
(2)是否存在正整数m，使得1/a1+1/a2+...+1/an>=1?若存在，求m的最小值；若不存在，说明理由

a(n)=aq^(n-1),
125 = a(1)a(2)a(3)=a^3q^[0+1+2]=a^3q^3, 5 = aq = a(2).
10 = |a(2) - a(3)|, a(3) = 15, 或，a(3) = -5.
a(3)=15时，q = a(3)/a(2)=3, a = 5/q = 5/3. a(n) = (5/3)3^(n-1) = 5*3^(n-2),
a(3)=-5时，q = a(3)/a(2)=-1, a= 5/q = -5, a(n) = -5(-1)^(n-1) = 5*(-1)^n.

a(n) = (5/3)3^(n-1)时，1/a(n) = (3/5)(1/3)^(n-1),
s(n)=1/a(1)+1/a(2)+...+1/a(n)
= (3/5)[1+1/3+...+(1/3)^(n-1)]
= (3/5)[1 - (1/3)^n]/(1-1/3)
= (9/10) - (9/10)(1/3)^n,
s(n)单调递增，9/10 > s(n) >= s(1) = 9/10 - 3/10 = 3/5.
不存在正整数m存在，使得 s(m) >= 1, 因为，对任意正整数m，都有 s(m) < 9/10 <1.

a(n) = -5(-1)^(n-1)时，1/a(n) = (-1/5)(-1)^(n-1),
s(n) = 1/a(1)+1/a(2)+...+1/a(n)
= (-1/5)[1 + (-1) + ... + (-1)^(n-1)]
= (-1/5)[1-(-1)^n]/[1-(-1)]
= (-1/10)[1-(-1)^n]
= -1/10 + (-1)^n/10,
s(2n-1) = -1/10 + (-1)^(2n-1)/10 = -1/5.
s(2n) = -1/10 + (-1)^(2n)/10 = 0.
同样，不存在正整数m，使得s(m) >= 1. 因为，对任意正整数m,都有s(m)<=0<1.