由S(n+1)/S(n)=(4n+2)/(n+1),可得a(n+1)/S(n) = S(n+1)/S(n) -1 = (3n+1)/(n+1),所以
S(n)=(n+1)/(3n+1) * a(n+1)
以n-1代替n可得S(n-1)=n/(3n-2) * a(n)
以上两式相减可得a(n)=(n+1)/(3n+1) * a(n+1) - n/(3n-2) * a(n)
所以a(n+1)/a(n) = 2(2n-1)/(n+1) * (3n+1)/(3n-2),
a(n)/a(n-1)= 2(2n-3)/n * (3n-2)/(3n-5),
.
a(2)/a(1) = 2/1 * 4/1
a(1) = 1.
以上各式相乘可得a(n)= 2^(n-1) * (2n-3)(2n-5)...3*1/(n!) * (3n-2)
= (2n-2)!/(n!* (n-1)!) * (3n-2).