Question

递增的等比数列﹛an﹜的前3项积为512，且这三项分别减去1,3,9后成等差数列，求证1/a1+2/a2+…﹢n/an﹤1

Answer 1

a1*a2*a3=512 =>(a1*q)^3=512 => a1*q=8 =>a2=8
(a1-1)+(a3-9)=2*(a2-3) => (a1-1)+(a2*q-9)=2*(8-3) => a1+8q=20 =>a2/q+8q=20

解得q=2 (q=1/2不合题意，舍去）
a1=4

an=a1*q^(n-1)=4*2^(n-1)=2^(n+1)

Bn=n/An=n/2^(n+1)

Sn=B1+b2+...+Bn
=1/4+2/8+...+(n-1)/2^n+n/2^(n+1)

2Sn=1/2+2/4+...+(n-1)/2^(n-1)+n/2^n

相减有

Sn=(1/2+1/4+1/8+...+1/2^n)-n/2^(n+1)
=(1/2)*[(1/2)^n-1]/(1/2-1)]-n/2^(n+1)
=1-(1/2)^n-n/2^(n+1)<1
得证