Pi in Pascal's Triangle (2014)

[senfiaj]

https://www.cut-the-knot.org/arithmetic/algebra/PiInPascal.shtml

[JohnKemeny]

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.

[dhosek → JohnKemeny]

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

[raldi]

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

[kevmo314 → raldi]

It's the equation in the diagram:

π = 3 + 2/3 (1/4 − 1/20 + 1/56 - ...)

[lixtra]

(2014) [1]

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

[nimonian]

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

[zkmon]

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

[dhosek → zkmon]

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).

[nimonian → dhosek]

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.

[hollerith → nimonian]

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

[dhosek → nimonian]

Oh, that’s really nice.

[madcaptenor → dhosek]

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

[dhosek → madcaptenor]

Depends on your starting point.

[adornKey → zkmon]

Indeed it does, e.g:

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