UC知道一条初中求值问题
来源：百度知道 编辑：UC知道 时间：2024/06/14 03:37:12
a1+1/a2=a2+1/a3=a3+1/a4=…=a（n-1）+1/an=an+1/a1，a（n-1）-an=1/a1-1/an，试推导（a1）平方（a2）平方…（an）平方的平方值是1。
请大虾写出详解！!!
a1-a2=1/a3-1/a2=(a2-a3)/(a2*a3)
a2*a3=(a2-a3)/(a1-a2)
a2-a3=1/a4-1/a3=(a3-a4)/(a3*a4)
a3*a4=(a3-a4)/(a2-a3)
2<=i<=n-1
a(i)*a(i+1)=(a(i)-a(i+1))/(a(i-1)-a(i))
a(n-1)-a(n)=1/a1-1/a(n)=(an-a1)/(a1*an)
a1*an=(an-a1)/(a(n-1)-a(n))
(a1*a2*...an)^2=a1*a2*(a2*a3)*(a3*a4)*...*(a(n-1)*an)*(an*a1)
=a1*a2*(a2-a3)/(a1-a2)*(a3-a4)/(a2-a3)*...*(a(n-1)-an)/(a(n-2)-a(n-1))*(an-a1)/(a(n-1)-an)
=a1*a2/(a1-a2)*(an-a1)
只能化简到这一步，是不是还有条件没写？
a1-a2=1/a3-1/a2=(a2-a3)/(a2*a3)
a2*a3=(a2-a3)/(a1-a2)
a2-a3=1/a4-1/a3=(a3-a4)/(a3*a4)
a3*a4=(a3-a4)/(a2-a3)
2<=i<=n-1
a(i)*a(i+1)=(a(i)-a(i+1))/(a(i-1)-a(i))
a(n-1)-a(n)=1/a1-1/a(n)=(an-a1)/(a1*an)
a1*an=(an-a1)/(a(n-1)-a(n))
(a1*a2*...an)^2=a1*a2*(a2*a3)*(a3*a4)*...*(a(n-1)*an)*(an*a1)
=a1*a2*(a2-a3)/(a1-a2)*(a3-a4)/(a2-a3)*...*(a(n-1)-an)/(a(n-2)-a(n-1))*(an-a1)/(a(n-1)-an)
=a1*a2/(a1-a2)*(an-a1)