Question

(1四次方+1/4)(3四次方+1/4)..(19四次方+1/4)/(2四次方+1/4)(4四次方+1/4)...(20四次方+1/4)

学霸速回，谢！！！(1四次方+1/4)(3四次方+1/4)..(19四次方+1/4)/(2四次方+1/4)(4四次方+1/4)...(20四次方+1/4)

Answer 1

答案应为 1/841

首先原式化为(4x1四次方+1)(4x3四次方+1)..(4x19四次方+1)/(4x2四次方+1)(4x4四次方+1)...(4x20四次方+1)

按照a^2+b^2=(a+b)^2-2ab
分解因式4x(2n-1)^4+1=[2x(2n-1)^2+1]^2-4x(2n-1)^2
=[2x(2n-1)^2+1+2x(2n-1)]x[2x(2n-1)^2+1-2x(2n-1)]
=(8n^2-4n+1)(8n^2-12n+5)

同理分解因式4x(2n)^4+1=(8n^2-4n+1)(8n^2+4n+1)
两式相除后消去相同项
所求s=s1s2s3....sn
sn=(8n^2-12n+5)/(8n^2+4n+1)
而8n^2-12n+5=(4n-3)^2+1
8n^2+4n+1=(4n+1)^2+1
s=(1^2+1)(5^2+1)(9^2+1)..(27^2+1)/(5^2+1)(9^2+1)(13^2+1)..(27^2+1)(41^2+1)
=(1^2+1)/(41^2+1)
=1/841
注：^2表示某个数的平方