圖解(所見及所得:What You See Is What You Get) $\frac{1}{3}$

n>1，$\frac{1+3+5...+(2n-1)}{(2n+1)+(2n+3)+(2n+5)+....+[2n+(2n-3)]+[2n+(2n-1)]}$ = $\frac{\frac{n[1+(2n-1)]}{2}}{\frac{n[(2n+1)+(4n-1)]}{2}}=\frac{2n}{6n}=\frac{1}{3}$