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=

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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=
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=

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)+…+1/(2n+2)
即:
an =1/(n+1)+1/(n+2)+1/(n+3)+…+1/2n ①式
a(n+1)= 1/(n+2)+1/(n+3)+…+1/2n+1/(2n+1)+1/(2n+2) ②式
对比①、②式,发现两式对齐的项相等,故:
a(n+1)-an=1/(2n+1)+1/(2n+2) - 1/(n+1)
为所求.