Theorem T8 13=1+35+7=1+3+57+9+11=⋯=1+3+⋯+(2n−1)(2n+1)+(2n+3)+⋯+(2n+2n−1)\begin{aligned} \frac{1}{3} &= \frac{1+3}{5+7} \\ &= \frac{1+3+5}{7+9+11} \\ &= \cdots \\ &= \frac{1+3+\cdots+(2n-1)}{(2n+1)+(2n+3)+\cdots+(2n+2n-1)} \end{aligned} 31=5+71+3=7+9+111+3+5=⋯=(2n+1)+(2n+3)+⋯+(2n+2n−1)1+3+⋯+(2n−1) for integer n≥1n\geq1n≥1.