Question

求和:(x+1/x)2(的平方)+(x2+1/x2)2+.(xn+1/xn)2

Answer 1

:(x+1/x)2(的平方)+(x2+1/x2)2+.(xn+1/xn)2

=(x^2+1/x^2+2)+(x^4+1/x^4+2)+(x^6+1/x^6+2)+.(x^2n+1/x^2n+2)

=(x^2+x^4+x^6+.+x^2n)+(1/x^2+1/x^4+.+1/x^2n)+2n

=x^2*(x^2n-1)/(x^2-1)+1/x^2*(1/x^2n-1)/(1/x^2-1)+2n

=x^2(x^2n-1)/(x^2-1)+1/x^2*(1-x^2n)/x^2n*x^2/(1-x^2)+2n

=x^2(x^2n-1)/(x^2-1)+(x^2n-1)/x^2n(x^2-1)+2n

=[x^(2+2n)(x^2n-1)+(x^2n-1)]/x^2n(x^2-1)+2n

=(x^(4n+2)-x^(2n++2)+x^2n-1)/x^2n(x^2-1)+2n

=