Pi in Pascal's Triangle (2014)

cut-the-knot.org

68 points·senfiaj··14 comments

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π - 2 = 1/1 + 1/3 - 1/6 - 1/10 + 1/15 + 1/21 - 1/28 •••

That's beautiful. I wonder why the -2 is there, though. To fix it, we would need

π = -x + 1/1 + 1/3 - 1/6 - 1/10 + 1/15 + 1/21 - 1/28 •••

where x = 2, and so it would be

π = -1/½ + 1/1 + 1/3 - 1/6 - 1/10 + 1/15 + 1/21 - 1/28 •••

which makes the -1st triangle number ½, I guess.

0th triangle number and I think that can probably be connected to 1+1-1+1-1+⋯=½ somehow.

Even after reading this, I don't understand what pi has to do with 4, 20, 56, 120, 220, 364

It's the equation in the diagram:

π = 3 + 2/3 (1/4 − 1/20 + 1/56 - ...)

(2014) [1]

[1] https://web.archive.org/web/20140901000000*/https://www.cut-...

There's a typo in the final line of math. I think 1/7C2 should be positive rather than negative.

I wonder if the triangle hides any secrets related to prime numbers as well.

It’s possible, I suppose.

One of my favorite proofs that the sums of each row are powers of two comes from the fact that the numbers in row n+1 are the coefficients of the powers of (a+b)ⁿ, so setting a=b=1 you get 2ⁿ (most discrete math students seeking to prove this end up reaching for induction which is a heavier proof than this).

I like the argument that every number in the row below is formed by summing two numbers from above. So each number above appears twice below. Hence the sum doubles.

You mean, every number in the upper row contributes twice to the lower row.

Oh, that’s really nice.

That requires you to prove the binomial theorem first, though, and won't that need induction?

Depends on your starting point.

Indeed it does, e.g:

https://nonagon.org/ExLibris/paul-erdos-bertrands-conjecture

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