# Pascal's hockey shot

**Discrete Mathematics**Level 2

\[ \begin{array}{c} 1 \end{array} \\ \begin{array}{cc} 1 & 1 \end{array} \\ \begin{array}{ccc} 1 & 2 & \color{red}{1}\end{array} \\ \begin{array}{cccc} 1 & 3 & \color{red}{3} & 1\end{array} \\ \begin{array}{ccccc} 1 & 4 & \color{red}{6} & 4 & 1\end{array} \\ \begin{array}{cccccc} \vdots & \hphantom{\vdots} & \vdots & \hphantom{\vdots} & \vdots \end{array}\\ \begin{array}{cccccc} 1 & 25 & \color{red}{300} & 2300 & 12650 & \cdots \end{array} \\ \begin{array}{ccccccc} 1 & 26 & 325 & 2600 & 14950 & \cdots & \hphantom{1000} \end{array} \\ \]

Pascal's Triangle is shown above for the \(0^\text{th}\) row through the \(4^\text{th}\) row, and part of the \(25^\text{th}\) and \(26^\text{th}\) rows are also shown above.

What is the sum of all the \(2^\text{nd}\) elements of each row up to the \(25^\text{th}\) row?

**Note**: The visible elements to be summed are highlighted in red.

**Additional clarification**: The topmost row in Pascal's Triangle is the \(0^\text{th}\) row. Then, the next row down is the \(1^\text{st}\) row, and so on. The leftmost element in each row of Pascal's Triangle is the \(0^\text{th}\) element. Then, the element to the right of that is the \(1^\text{st}\) element in that row, and so on.

**Your answer seems reasonable.**Find out if you're right!

**That seems reasonable.**Find out if you're right!

Already have an account? Log in here.