Question

急求，高手！数列an中,a1=0,a(n+1)=[(n+2)/n]an+1/n,求an的通项公式

Answer 1

两边同除以(n+1)(n+2)，得
a(n+1)/[(n+1)(n+2)]=an/[n(n+1)]+1/n(n+1)(n+2)
令，bn=an/[n(n+1)]
则,b(n+1)-bn=1/n(n+1)(n+2)
然后对等式右边进行拆项
b(n+1)-bn=1/n(n+1)(n+2)=[1/n(n+1)-1/(n+1)(n+2)]/2=[1/(n)-1/(n+1)-1/(n+1)+1/(n+2)]/2

然后用叠加法、
b2-b1=(1-1/2-1/2+1/3)/2
b3-b2=(1/2-1/3-1/3+1/4)/2
b4-b3=(1/3-1/4-1/4+1/5)/2
………
bn-b(n-1)=[1/(n-1)-1/(n)-1/(n)+1/(n+1)]/2
全部相加，得
bn-b1=[1-1/2-1/(n)+1/(n+1)]
又b1=a1/2=0。所以整理得
bn=1/4-1/[2n(n+1)]
所以an=n(n+1)/4-1/2