已知數列{An}的前n項和為Sn=1+m*An(m為不為1的常數），則通項公式An=______

熱心網友

Sn=1+m*AnSn-1=1+mAn-1Sn -Sn-1 =(1+m*An)-(1+mAn-1)An=m*An-mAn-1An/(mAn-1)=m/(m-1) (n=1)A1=1+m*A1A1=1/(1-m)An=[1/(1-m)]*{[m/(m-1)]^(n-1)}因為A1符合An所以An=[1/(1-m)]*{[m/(m-1)]^(n-1)} （n屬于N）

熱心網友

要過程嗎解：n=1, A1=S1=1+mA1 A1=1/(1-m)n=2An=Sn-S(n-1)An=1+mAn-(1+mA(n-1))An=m/(m-1)*A(n-1)則A(n-1)=m/(m-1)*A(n-2) A(n-2)=m/(m-1)*A(n-3)............類推下來A2=m/(m-1)*A1左邊，右邊分別相乘得Ａn=1/(1-m)*(m/(m-1))^(n-1)