已知X=1/2(5^1/n-5^-1/n),n∈N+,求(X+√1+x^2)^n的值

Question

已知X=1/2(5^1/n-5^-1/n),n∈N+,求(X+√1+x^2)^n的值

Answer 1

解 (X+√1+x^2)^n分子有理化得：[1/(-x+√1+x^2)]^n;
又把X=1/2(5^1/n-5^-1/n)带入1+x^2得：
1+x^2=1+1/4(5^1/n-5^-1/n)^2=1/4(5^1/n+5^-1/n)^2
所以
x+√1+x^2
=1/2(5^1/n-5^-1/n)+1/2(5^1/n+5^-1/n)
=5^-1/n
故：[1/(-x+√1+x^2)]^n=[1/（5^-1/n）]^n=5