# Difference between revisions of "2021 April MIMC 10 Problems/Problem 17"

The following expression $$\sum_{k=1}^{60} {60 \choose k}+\sum_{k=1}^{59} {59 \choose k}+\sum_{k=1}^{58} {58 \choose k}+\sum_{k=1}^{57} {57 \choose k}+\sum_{k=1}^{56} {56 \choose k}+\sum_{k=1}^{55} {55 \choose k}+\sum_{k=1}^{54} {54 \choose k}+...+\sum_{k=1}^{3} {3 \choose k}-2^{10}$$ can be expressed as $x^{y}-z$ which both $x$ and $y$ are relatively prime positive integers. Find $2^{x}(xy+2x+z)$.

$\textbf{(A)} ~4632 \qquad\textbf{(B)} ~4844 \qquad\textbf{(C)} ~4860\qquad\textbf{(D)} ~4864 \qquad\textbf{(E)} ~8960$

## Solution

To be Released on April 26th. $$\sum_{k=0}^{60} {60 \choose k}$$ can be expressed as $2^{60}$, and $60 \choose 0$ is equal to $1$. Therefore, we can simplify the original expression into $2^{60}-1+2^{59}-1+...+2^3-1-2^{10}=2^{60}+2^{59}+...+2^{3}+2^3-58-1024=2^{61}-(8+58+1024)=2^{61}-1090$. The expression that the answer wants would be $2^2\cdot(2\cdot 61+2\cdot2+1090)=4\cdot 1216=\fbox{\textbf{(D)} 4864}$.