A probability problem by D G

Let us consider a number $g$ , and try to find out $r_{\beta}(g)$ . First note that $f_{\beta}(r_\beta(g))=g\implies a_\beta+b_{\beta}=g$ . If $g\ge 2\beta$ , then, note that, since $b_{\beta}\le \beta$ , and $a_\beta\ge b_{\beta}$ , the only possible solution is $b_\beta=\beta,\ a_\beta=g-\beta$ . Then, we have $r_\beta(g)=\beta^{g-\beta}\cdot 1+\beta^{g-\beta-1}\cdot 0+\beta^{g-\beta-2}\cdot 1+\beta^{g-\beta-3}\cdot 2+\cdots+\beta^{0}\cdot (\beta-1)$ , i.e. $\kappa_\beta=\beta,\ \lambda_\beta=\beta^{g-\beta-2}\cdot 1+\beta^{g-\beta-3}\cdot 2+\cdots+\beta^{0}\cdot (\beta-1)$ . Thus, the required answer is $\sum_{i=2}^{\beta-1}i=\binom{\beta}{2}-1=\binom{100}{2}-1=\boxed{4949}$ .