已知数列满足a1=1,√an-1 - √an = √anan-1 ,求an=

A(n-1))^0.5-(An)^0.5=(An×A(n-1))^0.5
两边同除(An×A(n-1))^0.5
1/(An)^0.5-1/(A(n-1))^0.5=1
1/(A1)^0.5=1
数列{1/(An)^0.5}是以1为首项,1为公差的等差数列
1/(An)^0.5=n
An=1/n^2