拆项法求和S=（x+1/x)^+(x^+1/x^)^+.+（x*+1/x*)^*代表n次

答：
拆项法顾名思义就是把每项拆开.
S=(x+1/x)^2+(x^2+1/x^2)^2+...+(x^n+1/x^n)^2
=(x^2+1/x^2+2)+(x^4+1/x^4+2)+...+(x^(2n)+1/x^(2n)+2)
=(x^2+x^4+...+x^(2n))+(1/x^2+1/x^4+...+1/x^(2n))+(2+2+...+2)
=x^2(1-x^(2n))/(1-x^2)+1/x^2*(1-1/x^(2n))/(1-1/x^2)+2n
=x^2(1-x^(2n))/(1-x^2)+(1-1/x^(2n))/(x^2-1)+2n