Summing The Triangle

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 \begin{array}{c} 1 \end{array} \\ \begin{array}{cc} 1 & 1 \end{array} \\ \begin{array}{ccc} 1 & 2 & 1 \end{array} \\ \begin{array}{cccc} 1 & 3 & 3 & 1\end{array} \\ \begin{array}{ccccc} 1 & 4 & 6 & 4 & 1\end{array} \\ \begin{array}{cccccc} \vdots & \hphantom{\vdots} & \vdots & \hphantom{\vdots} & \vdots \end{array} \\

Pascal's triangle is shown above for the 0 th 0^\text{th} row through the 4 th 4^\text{th} row.

What is the sum of all the elements in the 1 2 th 12^\text{th} row?


Note : The topmost row in Pascal's triangle is the 0 th 0^\text{th} row. Then, the next row down is the 1 st 1^\text{st} row, and so on.


The answer is 4096.

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1 solution

Andy Hayes
Jun 27, 2016

Relevant wiki: Properties of Binomial Coefficients

If you sum all the elements in each row, you might notice a pattern:

Row Values in Row Sum 0 1 1 1 1 + 1 2 2 1 + 2 + 1 4 3 1 + 3 + 3 + 1 8 4 1 + 4 + 6 + 4 + 1 16 \begin{array}{c|c|c} \text{Row} & \text{Values in Row} & \text{Sum} \\ \hline 0 & 1 & 1 \\ 1 & 1+1 & 2 \\ 2 & 1+2+1 & 4 \\ 3 & 1+3+3+1 & 8 \\ 4 & 1+4+6+4+1 & 16 \\ \end{array}

These sums are all powers of 2 2 . In fact, the sum of all the elements in the n th n^\text{th} row is 2 n 2^n . This is an application of the identity:

k = 0 n ( n k ) = 2 n \sum\limits_{k=0}^{n}{\binom{n}{k}}=2^n

Thus, the sum of all the elements in the 1 2 th 12^\text{th} row is 2 12 = 4096 2^{12}=\boxed{4096} .

we can use the binomial theorem for this

abhishek alva - 4 years, 11 months ago

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