The Pascal’s Triangle can be derived from the powers of
two. This is how.
The leftmost column in the chart below shows, obviously, the powers of 2 in
ascending order.Each number in the second column is the sum of the number above and the number to the left.
Likewise each successive column.
Take the diagonal that rises upwards to the right from each
power of two.
The differences between the items in a diagonal equate to a
row of the Pascal’s Triangle. Examples
follow the chart.
0 0 0 0 0 0
1 1 1 1 1 1
2 3 4 5 6 7
4 7 11 16 22 29
8 15 26 42 64 93
16 31 57 99 163 256
32 63 120 219 382 638
64 127 247 466 848 1486
128 255 502 968 1816 3302
256 511 1013 1981 3797 7099
First Example:
The items in a diagonal that starts with 128 (2^7)
include: 128, 127, 120, 99, 64, and 29.
The differences between these items equate to row 7 in the
Pascal’s Triangle: 1, 7, 21, 35, 35 . .
. .
Second Example:
The items in a diagonal that starts with 256 (2^8)
include: 256, 255, 247, 219, 163, and 93.
The differences between these items equate to row 8 in the
Pascal’s Triangle: 1, 8, 28, 56, 70 . .
.
Just a reminder, the rows of the Pascal’s Triangle sum to
increasing powers of 2. And now you know
the reverse can happen: the powers of
two can produce the Pascal’s Triangle.
Since the diagonals in the Pascal’s Triangle sum to the
Fibonacci numbers, which approach the golden ratio, I guess it is acceptable to
say that you can find the golden ratio in powers of two, by this circuitous
route of finding the Pascal’s Triangle.
Posted on May 29, 2015
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