If $x_{k+1} = x_k + \frac12$ for $k=1, 2, \dots, n-1$ and $x_1=1$, find $x_1 + x_2 + \dots + x_n$. 
$\textbf{(A)}\ \frac{n+1}{2}\qquad\textbf{(B)}\ \frac{n+3}{2}\qquad\textbf{(C)}\ \frac{n^2-1}{2}\qquad\textbf{(D)}\ \frac{n^2+n}{4}\qquad\textbf{(E)}\ \frac{n^2+3n}{4}$
The sequence $x_1, x_2, \dots, x_n$ is an arithmetic sequence since every term is $\frac12$ more than the previous term. Letting $a=1$ and $r=\frac12$, we can rewrite the sequence as $a, a+r, \dots, a+(n-1)r$.
Recall that the sum of the first $n$ terms of an arithmetic sequence is $na+\binom{n}2r$.
Substituting our values for $a$ and $r$, we get $n+\frac{\binom{n}2}{2}$. Simplifying gives
$\boxed{\frac{n^2+3n}{4}\textbf{ (E)}}$