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=](/uploads/image/z/7759901-29-1.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%3D1%2Fn%2B1%2B1%2Fn%2B2%2B1%2Fn%2B3%2B%E2%80%A6%2B1%2F2n%EF%BC%88n%E2%88%88N%2B%29%2C%E5%88%99an%2B1-an%3D)
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.
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.
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