Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.2 - Page 257: 24

$\frac{k\times(k-1)}{2}$

. 1 + 2 + 3 + ··· + n = $\frac{n\times(n+1)}{2}$ Therefore, 1 + 2 + 3 + ··· + (k-1) = $\frac{(k-1)\times((k-1)+1)}{2}$ 1 + 2 + 3 + ··· + (k-1) = $\frac{(k-1)\times(k)}{2}$ Or, 1 + 2 + 3 + ··· + (k-1) = $\frac{k\times(k-1)}{2}$