n≥2时,
bn=Tn-T(n-1)=n²·bn-(n-1)²·b(n-1)
(n²-1)bn=(n-1)²b(n-1)
(n+1)(n-1)bn=(n-1)²b(n-1)
(n+1)bn=(n-1)b(n-1)
bn/b(n-1)=(n-1)/(n+1)
b(n-1)/b(n-2)=(n-2)/n
…………
b2/b1=1/3
连乘
bn/b1=(1/3)(2/4)...[(n-1)/(n+1)]=[1·2·...·(n-1)]/[3·4·...·(n+1)]=(1×2)/[n(n+1)]=2/[n(n+1)]
bn=2b1/[n(n+1)]=2/[n(n+1)]
数列{bn}的通项公式为bn=2/[n(n+1)]