求证:数列lg100,lg(100sinπ/4),lg(100sin^2 π/4),…,lg(100sin^n-1 π/
1个回答

an=lg(100sin^n-1 π/4)

an-1=lg(100sin^n-2 π/4)

an-an-1

=lg(100sin^n-1 π/4)-lg(100sin^n-2 π/4)

=lg[(100sin^n-1 π/4)/(100sin^n-2 π/4)]

=lg(sinπ/4)

=lg(√2/2)

所以数列是以lg100既以2为首项,公差为lg(√2/2)的等差数列.

an=2+(n-1)lg(√2/2)

设第n项小于0则有:

an=an=2+(n-1)lg(√2/2)0

得:

(4/lg2)+2>n >(4/lg2)+1

4/lg2≈13.3

所以

15.3>n>14.3

取整得 n=15

从第15项起其后每一项都是负数.