4.3.10 problem 49

Let $I_j$ be the indicator variable for the $j$-th prize being selected. The value recieved from the $j$-th prize is $j{I}_j$. Then, the total value $X$ is ${\sum }_{j=1}^n\left(j{I}_j\right)$. $\text{E}\left(X\right)={\sum }_{j=1}^n\mathrm{jP}(I_j=1)={\sum }_{j=1}^{n}j\frac{\left(\frac{n-1}{k-1}\right)}{\left(\frac{n}{k}\right)}={\sum }_{j=1}^{n}j\frac{k}{n}=\frac{k}{n}\frac{n(n+1)}{2}=\frac{k(n+1)}{2}.$