数列求和sn=(x+1/x)^2+(x^2+1/x^2)^2+……（x^n+1/x^n)^2

来源：学生作业帮助网编辑：六六作业网时间：2024/12/27 22:36:32

数列求和sn=(x+1/x)^2+(x^2+1/x^2)^2+……（x^n+1/x^n)^2数列求和sn=(x+1/x)^2+(x^2+1/x^2)^2+……（x^n+1/x^n)^2数列求和sn=(

数列求和sn=(x+1/x)^2+(x^2+1/x^2)^2+……（x^n+1/x^n)^2
数列求和sn=(x+1/x)^2+(x^2+1/x^2)^2+……（x^n+1/x^n)^2

数列求和sn=(x+1/x)^2+(x^2+1/x^2)^2+……（x^n+1/x^n)^2
当x=±1时,Sn=4n
当x≠±1时,
Sn=(x^2 +2 +1/x^2)+(1/x^4 +2 +1/x^4)+……+[x^(2n) +2 +1/x^(2n)]
=[x^2 +x^4 +……+x^(2n)] +2n +[1/x^2 +1/x^4 +……+1/x^(2n)]
=[x^2 -x^(2n+2)]/(1-x^2) +(1- 1/x^(2n))/(x^2 -1) +2n
=[x^(2n+2) -x^2 +1]/(x^2 -1) - 1/[x^(2n) (x^2 -1)] +2n