The sum of all numbers of the form $2k + 1$, where $k$ takes on integral values from $1$ to $n$ is: 
$\textbf{(A)}\ n^2\qquad\textbf{(B)}\ n(n+1)\qquad\textbf{(C)}\ n(n+2)\qquad\textbf{(D)}\ (n+1)^2\qquad\textbf{(E)}\ (n+1)(n+2)$
The sum of the odd integers $2k-1$ from $1$ to $n$ is $n^2$. However, in this problem, the sum is instead $2k+1$, starting at $3$ rather than $1$. To rewrite this, we note that $2k-1$ is $2$ less than $2k+1$ for every $k$ we add, so for $n$ $k$'s, we subtract $2n$, giving us $n^2+2n$,which factors as $n(n+2) \implies \boxed{\text{(C)} n(n+2)}$.