An=1/(n+1)+1/(n+2)+…+1/(2n-1)+1/(2n)
则
An+1=1/(n+2)+1/(n+3)+…+1/(2n-1)+1/(2n)+ 1/(2n+1)+1/(2n+2)
则
An+1-An
=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)-1/(2n+2)
=1/[(2n+1)(2n+2)]
设an=(1/n+1)+(1/n+2)+(1/n+3)+...+1/2n,则an+1-an等于?
1个回答
相关问题
-
设an=1+1/2+1/3+1/4+...+1/3n-1(n∈N*),则a(n+1)-an=
-
设数列满足an=1,a2=2,an=1/3(an-1+2an-2)(n∈N*且n≥3）,1.求证数列an+1-an是等比
-
数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),(1)设bn=2^n/an,
-
an+1=1/2 an^2-1/2 nan +1,a1=3,an=n+2,当n》2,an^n》an^n
-
在数列{an}中,an=1 1/2 1/3 … 1/3n-1(n属于N*),则an 1 -an=
-
设a0为常数，且an=3n-1-2an-1（n∈N*）．证明：n≥1时，an=[1/5][3n+（-1）n-1•2n]+
-
已知数列{an}中,a1=1,an=an-1·3^(n-1)(n≥2),设函数f(n)=log3an/9^n(n∈N*)
-
设数列{an}中，a1=a，an+1+2an=2n+1（n∈N*）．
-
数列{an}中,a1=1/2,an=an-1+1/n2+n(n≥2,n∈N+)则通项公式an
-
请问若数列{an}中的an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/2n(n∈N*).则an+1的表