Question

已知a(n+1)=an+1/n(n+2),求an的通项公式。

Answer 1

a(n+1)-an=1/n(n+2)=1/2[1/n-1/(n+2)]

an-a(n-1)=1/2[1/(n-1)-1/(n+1)]
……
a2-a1=1/2[1/1-1/3]
相加
an-a1=1/2*[1-1/3+1/2-1/4+……+1/(n-2)-1/n+1/(n-1)-1/(n+1)]
=1/2[1+1/2-1/n-1/(n+1)]
an=a1+3/4-(2n+1)/(2n²+2n)

Answer 2

a(n+1)=an+1/n(n+2),
a(n+1)-an=1/n(n+2)=(1/2)[1/n-1/(n+2)]
an-a(n-1)=(1/2)[1/(n-1)-1/(n+1)]
…………
a3-a2=(1/2)[1/2-1/4]
a2-a1=(1/2)[1-1/3]
所以
a(n+1)-a1=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
=(1/2)[3/2-(2n+3)/(n+1)(n+2)]
=3/4-(2n+3)/(n+1)(n+2)
没说a1是多少吗

Answer 3

设x=n+1则：a(x)=a(x-1)+1/(x-1)(x=1)
即：a(x)=a(x-1)+1/(x^2-1)