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在中学代数中常可见到这样的习题:已知x+x~(-1)=p,求x~2+x~(-2)、x~3+x~(-3)、x~4+x~(-4)等的值。通常的解法是 x~2+x~(-2)=(x+x~(-1))~2-2=p~2-2; x~3+x~(-3)=(x+x~(-1))~3-3(x+x~(-1))=p~3-3p; x~4+x~(-4)=(x~2+x~(-2))-2=(p~2-2)~2-2=p~4-4p~2+2 现在要问:一般情况下,已知x+x~(-1)=p,(x≠0,p∈R),x~n+x~(-n)(n∈Ⅳ)的值如何求?本文给出两种递归方法,介绍一个计算公式,并对其中一些情形进行讨论。
In the middle school algebra can often see such exercises: known x + x ~ (-1) = p, seeking x ~ 2 + x ~ (-2), x ~ 3 + x ~ (-3), x ~ 4 + Values such as x~(-4). The usual solution is x~2+x~(-2)=(x+x~(-1))~2-2=p~2-2; x~3+x~(-3)=(x+ x~(-1))~3-3(x+x~(-1))=p~3-3p; x~4+x~(-4)=(x~2+x~(-2) )-2=(p~2-2)~2-2=p~4-4p~2+2 Now to ask: Under normal circumstances, it is known that x+x~(-1)=p, (x≠0) , p ∈ R), how to find the value of x~n+x~(-n)(n ∈ IV)? In this paper, we give two recursive methods, introduce a calculation formula, and discuss some of them.