求和：Sn=1a+2a2+3a3+…+nan（a≠0）．

Question

求和：S_n=(1)/(a)+(2)/(a2)+(3)/(a3)+…+(n)/(an)（a≠0）．

Answer 1

当a=1时，Sn＝1+2+3+…+n＝(n(n+1))/(2)．
当a≠1时，S_n=(1)/(a)+(2)/(a2)+(3)/(a3)+…+(n)/(an)，①
(1)/(a)Sn=(1)/(a2)+(2)/(a3)+(3)/(a4)+…+(n)/(an+1)，②
①-②，得：(1−(1)/(a))Sn=(1)/(a)+(1)/(a2)+…+(1)/(an)−(n)/(an+1)
=((1)/(a)(1−(1)/(an)))/(1−(1)/(a))-(n)/(an+1)，
∴S_n=((1)/(a)(1−(1)/(an)))/((1−(1)/(a))2)−((n)/(an+1))/(1−(1)/(a))．
∴Sn＝

(n(n+1))/(2)，a＝1

((1)/(a)(1−(1)/(an)))/((1−(1)/(a))2)−((n)/(an+1))/(1−(1)/(a))，a≠1

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