Prove from the formula ⁿCr = (n!)/(r! (n - r)!) that ⁿCr + ⁿC_(r - 1) = ^(n + 1)Cr for 1 ≤ r ≤ n.

Step 1. Write the terms. $^{n}C_{r} = \dfrac{n!}{r!\,(n - r)!}$ and $^{n}C_{r - 1} = \dfrac{n!}{(r - 1)!\,(n - r + 1)!}$.

Step 2. A common bottom. Multiply the first by $\dfrac{n - r + 1}{n - r + 1}$ and the second by $\dfrac{r}{r}$. Both bottoms become $r!\,(n - r + 1)!$.

Step 3. Add the tops. $n!\,(n - r + 1) + n!\,r = n!\,(n + 1) = (n + 1)!$.

Step 4. Read it off. The sum is $\dfrac{(n + 1)!}{r!\,(n + 1 - r)!} = {}^{n + 1}C_{r}$.