jmc

By the arithmetic series formula, it follows that $$1_6+2_6+3_6+\cdots +45_6=\frac{45_6\times 50_6}{2}$$(notice that this formula remains the same as the base $10$ formula, as the derivation remains the same). We can ignore the $0$ for now, and evaluate the product $45_6\times 5_6$ (and append a $0$ at the end). Evaluating the units digit, we need to multiply $5_6\times 5_6=25_{10}=41_6$. Thus, the next digit is a $1$ and $4$ is carried over. The next digits are given by $4_6\times 5_6+4_6=24_{10}=40_6$. Writing this out: \begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c} & & & & & \stackrel{4}{4} & \stackrel{}{5}_6 \\ & & & \times & & 5 & 0_6 \\ \cline{4-7} & & & 4 & 0 & 1 & 0_6 \\ \end{array}Now, we divide by $2$ to obtain that the answer is ${\boxed{2003}}_6$. $$\begin{array}{ccccc}\text{multicolumn}2r2& 0& 0& 3& \\ \text{cline}2-52& 4& 0& 1& 0\\ \text{multicolumn}2r4& \downarrow & \downarrow & & \\ \text{cline}2-2\text{multicolumn}2r0& 0& 1& & \\ \text{multicolumn}2r& & 0& \downarrow & \\ \text{cline}4-4\text{multicolumn}2r& & 1& 0& \\ \text{multicolumn}2r& & 1& 0& \\ \text{cline}4-5\text{multicolumn}2r& & & 0& \end{array}$$We divide as we do normally; notice that $10_6÷2_6=3_6$.