Question

已知数列{an}满足a1=1/2,an=an-1+1/n^2-1(n>=2)，则{an}的通项公式为？

Answer 1

∵an=a(n-1)+1/(n^2-1)
∴an-a(n-1)=1/(n^2-1)=1/(n-1)(n+1)=1/2*[1/(n-1)-1/(n+1)]
即an-a(n-1)=1/2*[1/(n-1)-1/(n+1)]
a(n-1)-a(n-2)=1/2*[1/(n-2)-1/n]
…………………………………………
a2-a1=1/2*[1-1/3]
累加得：an-a1=1/2*[1/(n-1)-1/(n+1)+1/(n-2)-1/n+……+1-1/3]
=1/2*[-1/(n+1)-1/n+1+1/2]
=3/4-(2n+1)/2n(n+1)
所以an=5/4-(2n+1)/2n(n+1)

Answer 2

an=5/4-(2n+1)/2n(n+1)