Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is
$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$
When $m=1$, $a_{n+1}=1+a_n+n$. Hence,
\[a_{2}=1+a_1+1\]
\[a_{3}=1+a_2+2\]
\[a_{4}=1+a_3+3\]
\[\dots\]
\[a_{12}=1+a_{11}+11\]
Adding these equations up, we have that $a_{12}=12+(1+2+3+...+11)=\boxed{\mathrm{(D) \ } 78}$
~AopsUser101