Question

已知数列｛an｝的通项公式为an=log2（n+1/n+2）(n∈N*),设其前n项和为Sn，则使Sn<-5成立的自然数n

Answer 1

a1=log2 2/3 a2=log2 3/4.....

a1+a2+...an=log2 (2/3)(3/4)(4/5)...(n+1/n+2)

=log2 2/(n+2)=Sn

sn<-5 即 2/(n+2)<2^(-5)

算得 n>62

n=63,64......

追问

为什麼sn<-5 会变成 2/(n+2)<2^(-5)

追答

log2 (2/(n+2))<-5

2/(n+2)<2^(-5)

Answer 2

Sn=a1+a2+......+an =log2(2/3)+log2(3/4)+.....+log2[(n+1)/(n+2)]
=log2[2/3×3/4×......×(n+1)/(n+2)]
=log2[2/(n+2)]=1-log2(n+2)<-5
∴log2(n+2)>6 ∴n+2>64 ∴ n>62且n为正整数
∴n最小63