116问答网 > x-1-x(x-1)+x(x-1)^2......-x(x-1)^2003+x(x-1)^2004

x-1-x(x-1)+x(x-1)^2......-x(x-1)^2003+x(x-1)^2004

2025-06-28 01:16:31

推荐回答（1个）

回答1：

x-1-x(x-1)+x(x-1)^2......-x(x-1)^2003+x(x-1)^2004
=x-1+{-x(x-1)+x(x-1)^2......-x(x-1)^2003+x(x-1)^2004}
=x-1+x{∑(-1)^n*(x-1)^n} [n=1～2004]

a[n]=(-1)^n*(x-1)^n为等比数列
a[1]=-(x-1),q=-(x-1)
S{a[n]}=a1(1-q^n)/(1-q)
=(1-x)(1-(1-x)^n)/{1+(x-1)}
={1-x-(1-x)^n}/x

所以：
x-1-x(x-1)+x(x-1)^2......-x(x-1)^2003+x(x-1)^2004
=x-1+x{∑(-1)^n*(x-1)^n} [n=1～2004]
=x-1+{1-x-(1-x)^2004}
=-(1-x)^2004