圖 示 $\frac{1}{3}=\frac{1+3}{5+7}=\frac{1+3+5}{7+9+11}$

等差級數有n項，首項2n+1，公差2，則

(2n+1)+(2n+3)+(2n+5)+..........+[2n+(2n-1)]

=n‧2n+[1+3+5+......(2n-1)]=n‧2n+n²=3n²

=3[1+3+5+......(2n-1)]

所以 $\small\displaystyle\frac{1+3+5+...............+(2n-1)}{(2n+1)+(2n+3)+(2n+5)+..........+[2n+(2n-1)]}=\frac{1}{3}$

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