1)自己算
2)可以猜,也可算出
a1+a2+.+an=(2n-1)nan
a1+a2+.+a(n+1)=(2n+1)(n+1)a(n +1)
a(n+1)=(2n+1)(n+1)a(n +1)-(2n-1)nan
(2n+3)a(n+1)=(2n-1)an
a(n+1)/an=(2n-1)/(2n+3)
a2/a1=1/5
a3/a2=3/7
.
an/a(n-1)=(2n-3)/(2n+1)
an/a1=3/(4n^2-1)
an=1/[(2n-1)(2n+1)]
3）當(dāng)n=1
a1=1/3
當(dāng)n=k
ak=1/[(2k-1)(2k+1)]
當(dāng)n=k+1
由a1+a2+.+an=(2n-1)nan
a1+a2+.+a(n+1)=(2n+1)(n+1)a(n +1)
得 (2k+3)a(k+1)=(2k-1)ak
a(K+1)=ak(2K-1)/(2k+3)
a(k+1)=1/[(2k+1)(2K+3)]
成立