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)].