一道高中竞赛问题

解 我们有更一般结论为:T(2n)=a(2)*a(4)*...*a(2n)/[a(1)*a(3)*...*a(2n-1)]其中an=n^4+1/4.a(n)=n^4+1/4=(n^2-n+1/2)*(n^2+n+1/2).令b(n)=n^2-n+1/2, c(n)=n^2+n+1/2.易验证:b(2n)=4n^2-2n+1/2=(2n-1)^2+(2n-1)+1/2=c(2n-1), c(2n)=4n^2+2n+1/2=(2n+1)^2-(2n+1)+1/2=b(2n+1).b(n)*c(n)=(n^2+1/2)^2-n^2=n^4+1/4=a(n)，所以 T(2n)=b(2)c(2)*b(4)c(4)...(b(2n)c(2n)/[b(1)c(1)*b(3)*c(3)...(b(2n-1)c(2n-1)]即 T(2n)=c(2n)/b1=8n^2+4n+1.于是 T(10)=c(10)/b(1)=(10^2+10+1/2)/(1/2)=221. 收起