an=Sn-S(n-1)=n^2an-(n-1)^2a(n-1)
整理得an/a(n-1)=(n-1)/(n+1)
累乘得an=2/[n(n+1)]其实我就是想问下那个累乘的过程.....我知道an/a(n-1)=n-1/n+1,an/a1=an/a(n-1)*a(n-1)/a(n-2)*......*a2/a1,然后怎么算呢？an/a(n-1)=n-1/n+1an/a1=an/a(n-1)*a(n-1)/a(n-2)*.....*a2/a1倒过来写更方便看些：an/a1=a2/a1*a3/a2*a4/a3*……*an/a(n-1)=(1/3)*(2/4)*(3/5)*(4/6)*……*(n-1)/(n+1)=(1乘到n-1)/(3乘到n+1)其中3乘到n-1被约掉得2/[n(n+1)]