TeX source:
\begin{align*} 1 + 2 + ... + 99 + 100 &= (1 + 2 + ... + 99) + 100\\ &= \frac{99 \cdot (99 + 1)}{2} + 100 \; \; \; \mathrm{by \; our \; assumption \; of \; p(99)} \\ &= \frac{99 \cdot 100}{2} + \frac{2 \cdot 100}{2}\\ &= \frac{100 \cdot 101}{2} \\ &= \frac{100 \cdot (100 + 1)}{2} \end{align*}