当n=1时,A1=1/2An=1/(1×2)=1/2原式成立设当n=k时,Ak=1/(k(k+1))则n=k+1时A1+A2+……+Ak+A(k+1)=(k+1)^2×A(k+1)又∵A1+A2+……+Ak=k^2×Ak=k^2×1/(k(k+1))=k/(k+1)两式相减A(k+1)=(k+1)^2×A(k+1)-k/(k+1)(k^2+2k)A(k+1)=k/(k+1)A(k+1)=1/((k+1)(k+2))原式也成立。An=1/(n(n+1))