Question

数学级数,问题!在求幂级数的和函数时,书上有如下等式“1/(1-x)=1+x+x^2+…+x^n+…(-1

Answer 1

等比数列求和

Sn=a1(1-q^n)/(1-q)=(a1-anq)/(1-q)

x^0+x^1+x^2+...x^n+…是 x的等比数列

x^0+x^1+x^2+...x^n+…=（a1-anq)/(1-q)

=（1-(x^n)x)/1-x 因为-1