Chapter 15 - Recursion - Section 15.2 - Case Study: Computing Factorials - Check Point - MyProgrammingLab - Page 503: 15.5

Let $f(n) = 1+2+3+...+n,$ then the recursive definition will be $f(1) = 1;$ $f(n) = n+f(n-1); n>=1$

(1) $f(n) = 1+2+3+...+n-1+n $ Also (2) $f(n-1) = 1+2+3+...+n-1$ Replacing $1+2+3+...+n-1$ by $f(n-1)$ from equation 2 in equation 1 we get $f(n) = n+f(n-1); n>=1$