1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 ⋮ ⋮ ⋮ ⋮ ⋮
Pascal's triangle is shown above for the 0 th row through the 4 th row.
What is the sum of all the elements in the 1 2 th row?
Note
: The topmost row in Pascal's triangle is the
0
th
row. Then, the next row down is the
1
st
row, and so on.
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Relevant wiki: Properties of Binomial Coefficients
If you sum all the elements in each row, you might notice a pattern:
Row 0 1 2 3 4 Values in Row 1 1 + 1 1 + 2 + 1 1 + 3 + 3 + 1 1 + 4 + 6 + 4 + 1 Sum 1 2 4 8 1 6
These sums are all powers of 2 . In fact, the sum of all the elements in the n th row is 2 n . This is an application of the identity:
k = 0 ∑ n ( k n ) = 2 n
Thus, the sum of all the elements in the 1 2 th row is 2 1 2 = 4 0 9 6 .