1/3+1/(3+6)+1/(3+6+9)+……+1/(3+6+9+……+99)
首先,提出1/3得到:1/3*[1+1/(1+2)+1/(1+2+3)+…+1/(1+2+3+…+33)]
然后利用等差数列求和公式得到通项公式为:n(n+1)/2则
1/3*[1+1/(1+2)+1/(1+2+3)+…+1/(1+2+3+…+33)]
=1/3*{2/[1(1+1)]+2/[2(2+1)]+2/[3(3+1)]+…+2/[33(33+1)]}
=2/3*{1-1/2+1/2-1/3+…+1/33-1/34}
=2/3*{1-1/34}
=11/17