Test and analysis: From the nature of N(x), we can know that when x is odd, the largest odd factor of x is obviously itself. So N(x)=x, so we can decompose and calculate the sum of odd and even terms respectively, that is,

so = n( 1)+n(3)+…+n()= 1+3+…+=

When x is an even number and x∈[]

(1) When k= 1, x ∈ [2 2,4], the interval only contains two even numbers, and N(2)= 1, so the sum of the maximum odd factors of all even numbers in the interval is;

(2) When k=2, x ∈ [4 4,8], the interval contains even numbers of 4,6, so the sum of all the largest odd factor even numbers in this interval is

(3) When k=3, x∈[ 8, 16], and the interval contains an even number of 8, 10. , 12, 14, then the sum of the largest odd factors of all even numbers in the interval is, so we can draw that when x ∈.

∴ Sum k from 1 to n- 1.

,

To sum up:

Comments: This topic mainly examines the summation of series, and examines students' ability to analyze and solve problems through known conditions.