数列{an}的通项公式an=1/n+1+1/n+2+1/n+3+…+1/2n（n∈N+),则an+1-an=

答案：a(n+1)-an=1/(2n+1)+1/(2n+2) - 1/(n+1)因为an=1/(n+1)+1/(n+2)+1/(n+3)+…+1/2n所以求a(n+1)等于多少只要将(n+1)替代上式的n即可故：a(n+1)=1/(n+2)+1/(n+3)+1/(n+4)+…+1/(2（n+1）)=1/(n+2)+1/(n+3)+1/(n+4...