HMMT November

We have that $2^{3} \equiv 1(\bmod 7)$. Hence, it suffices to consider the exponents modulo 3. We note that the exponents are the triangular number and upon division by 3 give the pattern of remainders $1,0,0,1,0,0, \ldots$, so what we want is $$\begin{aligned} 2^{\frac{1 \cdot 2}{2}}+\cdots+2^{\frac{2011 \cdot 2012}{2}} & \equiv 2^{1}+2^{0}+2^{0}+2^{1}+\ldots+2^{0}+2^{1} \quad(\bmod 7) \\ & \equiv \frac{2010}{3}\left(2^{1}+2^{0}+2^{0}\right)+2^{1} \\ & \equiv(670)(4)+2 \\ & \equiv 1 \end{aligned}$$