Let $W_1 < W_2 < · · · < W_n$ be the order statistics of $n$ independent observations from a $U(0, 1)$ distribution. $(a)$ Show that $E((W_r)^2 ) = r(r + 1)/(n + 1)(n + 2)$
My attempt is displayed below, and I am not sure how to get the desired result. Perhaps I have made a mistake. Any help much appreciated.
Your integration step is not correct, as you have made an algebraic error by not expanding the binomial term $(1-u)^{r+1}$. But to do this explicitly is tedious and not a good idea. Instead, you should observe that the order statistic has PDF $$f_{W_r}(w) = \frac{n!}{(r-1)! (n-r)!} w^{r-1} (1-w)^{n-r} \, \quad 0 \le w \le 1,$$ which is also the PDF of a $\operatorname{Beta}(r,n-r+1)$ distribution. That is to say, we can take as a given that $$\int_{w=0}^1 f_{W_r}(w) \, dw = 1$$ for any positive integer choices $r \le n$. So when we calculate $w^2 f_{W_r}(w)$, we get up to a scaling factor, a $\operatorname{Beta}(r+2, n-r+1)$ distribution. Specifically, $$\begin{align*} \operatorname{E}[W_r^2] &= \int_{w=0}^1 \frac{n!}{(r-1)! (n-r)!} w^{r+1} (1-w)^{n-r} \, dw \\ &= \frac{(r+1)r}{(n+2)(n+1)} \int_{w=0}^1 \frac{(n+2)!}{(r+1)! (n-r)!} w^{r+1} (1-w)^{n-r} \, dw \\ &= \frac{(r+1)r}{(n+2)(n+1)}, \end{align*}$$ where the integral equals $1$ because this is the integral of the aforementioned $\operatorname{Beta}(r+2, n-r+1)$ PDF.