Saudi Arabia Mathematical Competitions

We have $$s_m=\sum _{k=0}^m\left(2^k-k\right)=2^{m+1}-1-\frac{m(m+1)}{2}$$ We prove that for $m\ge 3$, we have $2^m<s_m<2^{m+1}$. This inequality is equivalent to $$2^m<2^{m+1}-1-\frac{m(m+1)}{2}<2^{m+1}$$ The right inequality is obvious. The left inequality is equivalent to $$1+\frac{m(m+1)}{2}<2^m$$ that is $$2+2\frac{m(m+1)}{2}<2^{m+1}$$ or $$2\left(\left(\genfrac{}{}{0ex}{}{m+1}{0}\right)+\left(\genfrac{}{}{0ex}{}{m+1}{2}\right)\right)<2^{m+1}$$ and we are done.

For $m=0$, we have $s_0=1=2^0$. For $m=1$, we have $s_1=2=2^1$. For $m=2$, we have $s_2=4=2^2$. The solutions are $m\in \{0,1,2\}$.