在数列an中已知a1=1，且满足an+1-an=an⼀n+1,求通项公式

∵a(n+1)-an=an/(n+1)

∴a(n+1)=an*(n+2)/(n+1)
∴a(n+1)/an=(n+2)/(n+1)
那么an/a(n-1)=(n+1)/n
a(n-1)/a(n-2)=n/(n-1)
……………………………
a3/a2=4/3
a2/a1=3/2
累乘，得：an/a1=(n+1)/2
而a1=1，所以an=(n+1)/2

移项an到右边整理得（an+1）/(n+2)=an/(n+1)即有[（an+1）/(n+2)]/[an/(n+1)]=1恒成立即有数列{an/(n+1)}为首项为a1/2=1/2 公比为1的等比数列即有an/(n+1)=a1/2*1^(n-1)=1/2即有an=(n+1)/2

a(n+1)=(n+2)/(n+1)*an
a(n+1)/(n+2)=an/(n+1)
所以就是an/(n+1)=a(n-1)/n=a(n-2)/(n-1)=……=a1/(1+1)=1/2
所以 an=(n+1)/2

an+1=(n+2)an/(n+1);
an+1/an=(n+2)/(n+1);
an=an/an-1*(an-1/an-2)*...*(a3/a2)*(a2/a1)*a1
=(n+1)/n*(n/(n-1))*...*(4/3)*(3/2)*1
=(n+1)/2;