数列113，219，3127，4181，…的前n项和是n2+n+12−12•3nn2+n+12−12•3n．

Question

数列1(1)/(3)，2(1)/(9)，3(1)/(27)，4(1)/(81)，…的前n项和是

(n2+n+1)/(2)−(1)/(2•3n)

(n2+n+1)/(2)−(1)/(2•3n)

．

Answer 1

∵1(1)/(3)+2(1)/(9)+3(1)/(27)+4(1)/(81)+n(1)/(3n)=（1+2+3+…+n）+（(1)/(3)+(1)/(9)+…+(1)/(3n)）
=(n(n+1))/(2)+((1)/(3)−(1)/(3n+1))/(1−(1)/(3))=(n2+n+1)/(2)−(1)/(2•3n)
故答案为：(n2+n+1)/(2)−(1)/(2•3n)