已知数列{an}中,a1=1/2,Sn=n^2*an(n属于N*)

1.
a1=1/2
a2=S2-a1=4a2-1/2,3a2=1/2 a2=1/6
a3=S3-a2-a1=9a3-1/6-1/2 8a3=2/3 a3=1/12
a4=S4-a3-a2-a1=16a4-1/12-1/6-1/2 a4=1/20
2、
a1=1/2=1/(1×2) a2=1/6=1/(2×3) a3=1/12=1/(3×4) a4=1/20=1/(4×5)
猜想：an=1/[n(n+1)]
证：
n=1时,a1=1/2=1/(1×2),等式成立.
假设当n=k (k∈N+)时,等式成立,即ak=1/[k(k+1)].则当n=k+1时,
Sk+1=(k+1)²a(k+1)
Sk=k²ak
a(k+1)=Sk+1-Sk=(k+1)²a(k+1)-k²ak
(k²+2k)a(k+1)=k²ak
a(k+1)=k²ak/(k²+2k)
=k²/[k(k+1)(k²+2k)]
=k²/k²(k+1)(k+2)]
=1/[(k+1)(k+2)]
=1/[(k+1)(k+1+1)]
等式同样成立.
综上,得an=1/[n(n+1)].

1，a1+a2=S2=2²*a2；
a2=1/6;
a1+a2+a3=S3=9*a3;
8a3=2/3;
a3=1/12;
an=1/n(n+1);
2、Sn+1-Sn=(n+1)²an+1-n²an=an+1;
(n+1+1)(n+1-1)an+1=n²an;
(n+2)an+1=nan;
an+1=nan/(n+2);
a2=1*a1/3=1/6;
a3=2*a2/4=1/12;
a4=3*a3/5=1/20;
......
an=1/n(n+1)=1/n-1/(n+1);