已知 x + 1 x = 5 {\displaystyle x+{\frac {1}{x}}=5} ，求 x 10 + 1 x 10 = {\displaystyle x^{10}+{\frac {1}{x^{10}}}=} ? 我自己算是6375623，不知對不對？
x 10 + x − 10 {\displaystyle x^{10}+x^{-10}} = ( x 5 + x − 5 ) 2 − 2 {\displaystyle =(x^{5}+x^{-5})^{2}-2} x 5 + x − 5 {\displaystyle x^{5}+x^{-5}} = ( x + 1 / x ) 5 − 5 x 3 − 10 x − 10 / x − 5 x − 3 {\displaystyle =(x+1/x)^{5}-5x^{3}-10x-10/x-5x^{-3}} = ( x + 1 / x ) 5 − 5 ( ( x + 1 / x ) 3 − 3 x − 3 / x ) − 10 ( x + 1 / x ) {\displaystyle =(x+1/x)^{5}-5((x+1/x)^{3}-3x-3/x)-10(x+1/x)} = 5 5 − 5 ∗ 5 3 + 5 ∗ 5 {\displaystyle =5^{5}-5*5^{3}+5*5} = 2525 {\displaystyle =2525} ( x 5 + x − 5 ) 2 − 2 {\displaystyle (x^{5}+x^{-5})^{2}-2} = 2525 2 − 2 {\displaystyle =2525^{2}-2} = 6375623 {\displaystyle =6375623}
不一定要用二項式定理展開吧？ x 2 + x − 2 = ( x + x − 1 ) 2 − 2 ⋅ x ⋅ x − 1 = 5 2 − 2 = 23 {\displaystyle x^{2}+x^{-2}=(x+x^{-1})^{2}-2\cdot x\cdot x^{-1}=5^{2}-2=23} x 3 + x − 3 = ( x + x − 1 ) 3 − 3 ⋅ x ⋅ x − 1 ( x + x − 1 ) = 5 3 − 3 ⋅ 5 = 110 {\displaystyle x^{3}+x^{-3}=(x+x^{-1})^{3}-3\cdot x\cdot x^{-1}(x+x^{-1})=5^{3}-3\cdot 5=110} x 5 + x − 5 = ( x 2 + x − 2 ) ⋅ ( x 3 + x − 3 ) − ( x + x − 1 ) = 23 ⋅ 210 − 5 = 2525 {\displaystyle x^{5}+x^{-5}=(x^{2}+x^{-2})\cdot (x^{3}+x^{-3})-(x+x^{-1})=23\cdot 210-5=2525} x 10 + x − 10 = ( x 5 + x − 5 ) 2 − 2 = 2525 2 − 2 = 6375623 {\displaystyle x^{10}+x^{-10}=(x^{5}+x^{-5})^{2}-2=2525^{2}-2=6375623}
這類問題有個方法一層層遞推下去。 令 x k = x k + x − k {\displaystyle x_{k}=x^{k}+x^{-k}} ，則 x 1 = x 1 {\displaystyle x_{1}=x_{1}} x 2 = ( x 1 ) 2 − 2 {\displaystyle x_{2}=(x_{1})^{2}-2} x 3 = x 1 ⋅ x 2 − x 1 {\displaystyle x_{3}=x_{1}\cdot x_{2}-x_{1}} x 4 = ( x 2 ) 2 − 2 {\displaystyle x_{4}=(x_{2})^{2}-2} x 5 = x 2 ⋅ x 3 − x 1 {\displaystyle x_{5}=x_{2}\cdot x_{3}-x_{1}} x 6 = ( x 3 ) 2 − 2 {\displaystyle x_{6}=(x_{3})^{2}-2} x 7 = x 3 ⋅ x 4 − x 1 {\displaystyle x_{7}=x_{3}\cdot x_{4}-x_{1}} ........
