Question

已知数列{an}中，a1=1，a(n+1)=(1+2/n)an+1，求{an}的通项公式

Answer 1

解：
a(n+1)=(1+ 2/n)an +1=[(n+2)/n]an +1
a(n+2)=[(n+3)/(n+1)]a(n+1)+1
a(n+2)-a(n+1)=[(n+3)/(n+1)]a(n+1) -[(n+2)/n]an
a(n+2)=[2(n+2)/(n+1)]a(n+1) -[(n+2)/n]an
a(n+2)/(n+2)=2a(n+1)/(n+1) - an/n
a(n+2)/(n+2)-a(n+1)/(n+1)=a(n+1)/(n+1)-an/n
数列{an/n}是等差数列
a1/1=1/1=1
a2=(1+ 2/1)a1 +1=3a1+1=3×1+1=4
a2/2=4/2=2
a2/2 -a1/1=2-1=1
数列{an/n}是以1为首项，1为公差的等差数列
an/n=1+1×(n-1)=n
an=n²
数列{an}的通项公式为an=n²