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Pi in Pascal's Triangle (2014)

$\pi$ in Pascal's Triangle Daniel Hardisky discovered $pi$ in Pascal's Triangle - something every one was looking for. $\displaystyle\pi = 3+\frac{2}{3}\bigg(\frac{1}{C^{4}_{3}}-\frac{1}{C^{6}_{3}}+\frac{1}{C^{8}_{3}}-\cdot\bigg).$ This is Daniel's modification of the famous Nilakantha Somayaji (1444-1544) series $\begin{align}\displaystyle \pi &=3+\frac{4}{2\cdot 3\cdot 4}-\frac{4}{4\cdot 5\cdot 6}+\frac{4}{6\cdot 7\cdot 8}-\ldots\\ &=3+\frac{4}{6}\bigg(\frac{1\cdot 2\cdot 3}{2\cdot 3\cdot 4}-\frac{1\cdot 2\cdot 3}{4\cdot 5\cdot 6}+\frac{1\cdot 2\cdot 3}{6\cdot 7\cdot 8}-\ldots\bigg)\\ &=3+\frac{2}{3}\bigg(\frac{1}{C^{4}_{3}}-\frac{1}{C^{6}_{3}}+\frac{1}{C^{8}_{3}}-\ldots\bigg) \end{align}$ Daniel based his discovery on Tony Foster's observation that each of the denominators in Nilakantha's series is the area of a Pythagorean triangle.

Daniel based his discovery on Tony Foster's observation that each of the denominators in Nilakantha's series is the area of a Pythagorean triangle. With so defined $a,b,c,$ it is easy to see that $a^{2}+b^{2}=c^{2}.$ The area of the corresponding triangle is, say, $A=ab/2=mn(m^{2}-n^{2}),$ which for $n=1$ gives $A=(m-1)m(m+1),$ such that the denominators in the series result for $m=5,7,9,\ldots$ Of course, the Leibniz series $\displaystyle\frac{\pi}{4}=\sum_{n=1}(-1)^{n+1}\frac{1}{2n-1}$ can also be looked at as the sum of the reciprocals of the binomial coefficients:

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