已知数列{an}满足a1=1/2,a1+a2+……+an=n^2an(n∈N*),试用数学归纳法证明:an=1/[n(n+1)]

当n=1时显然成立.
若当n=m时也成立(m>=1),则当n=m+1时有
m^2am+a(m+1)=(m+1)^2a(m+1) =>m^2/[m(m+1)]=(m^2+2m)a(m+1)
=>a(m+1)=m^2/[(m^2+m)(m^2+2m)]
=>a(m+1)=1/[(m+1)(m+2)]
=>a(m+1)=1/{(m+1)[(m+1)+1)]}
得证!