求和Sn=(x+1/x)²+(x²+1/x²)²+...(x的n次方+1/x的n次方)&s
问题描述：
求和Sn=(x+1/x)²+(x²+1/x²)²+...(x的n次方+1/x的n次方)&s
每项都是平方
答
[x+(1/x)] ^2+[x^2+(1/x)]^2+...+[x^n+(1/x)]^2
=[x^2+x^4+x^6+ …… +x^(2n)]+2[1+x+x^2+ …… +x^(n-1)]+[x^-2+x^-4+x^-6+ …… +x^(-2n)]
=x^2[1-x^(2n)]/(1-x^2)+2[1-x^(n-1)]/(1-x)+x^(-2)[1-x^(-2n)]/[1-x(-2)]
=x^2[1-x^(2n)]/(1-x^2)+2[1-x^(n-1)](1+x)/(1-x^2)+[x^(-2n)-1]/(1-x^2)
={x^2[1-x^(2n)]+2[1-x^(n-1)](1+x)+[x^(-2n)-1]}/(1-x^2)
={x^2-x^(2n+2)+[2-2x^(n-1)](1+x)+x^(-2n)-1}/(1-x^2)
=[x^2-x^(2n+2)-2x^(n-1)+2x-2x^n+x^(-2n)+1]/(1-x^2)