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In 2007 J.C. Toloza discovered a new formula for $\pi$:

$\textstyle2+{2\choose2}^{-1}+{3\choose2}^{-1}-{4\choose2}^{-1}-{5\choose2}^{-1}+{6\choose2}^{-1}+{7\choose2}^{-1}-\cdots=\pi$

Toloza's formula essentially relates $\pi$ to one diagonal of Pascal's triangle via a pairwise alternating sum of reciprocals of binomial coefficients.

But Toloza's formula can be generalised to relate $\pi$ to every second diagonal of Pascal's triangle.

$\text{Suppose }T_p:=\textstyle\frac{2p}{2p-1}+\sum_{n=1}^\infty\sum_{r=0}^{2p-1}(-1)^{n-1}{2pn+r\choose 2p}^{-1}\text{ for positive integers }p.$

Then Toloza's formula is now $T_1=\pi$.

See if you can prove that for $p>1$

$T_p=\textstyle2\pi\sum_{i=1}^{p-1}(-1)^{i-1}{2p-2\choose i-1}(\csc\frac{\pi i}{2p}-1).$

For example, with $p=2$ we find the 4-fold alternating sum

$\textstyle\frac{4}{3}+{4\choose4}^{-1}+{5\choose4}^{-1}+{6\choose4}^{-1}+{7\choose4}^{-1}-{8\choose4}^{-1}-{9\choose4}^{-1}-{10\choose4}^{-1}-{11\choose4}^{-1}+\cdots=2(\sqrt2-1)\pi$

or with $p=3$, the 6-fold alternating sum

$\tiny{\textstyle\frac{6}{5}+{6\choose6}^{-1}+{7\choose6}^{-1}+{8\choose6}^{-1}+{9\choose6}^{-1}+{10\choose6}^{-1}+{11\choose6}^{-1}-{12\choose6}^{-1}-{13\choose6}^{-1}-{14\choose6}^{-1}-{15\choose6}^{-1}-{16\choose6}^{-1}-{17\choose6}^{-1}+\dots=2(5-\frac{8}{3}\sqrt3)\pi}$

etc.

tywebb
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  • Are you sure you want to use \tiny and \scriptsize? Instead, you can put $$ on both sides of the MathJax to make it larger and use \begin{align}\end{align} or \begin{aligned}\end{aligned} to align the MathJax – Тyma Gaidash Jun 10 '23 at 12:49
  • Can you make it clear what the question is? – Jakobian Jun 10 '23 at 12:52
  • Just to prove $T_p=\textstyle2\pi\sum_{i=1}^{p-1}(-1)^{i-1}{2p-2\choose i-1}(\csc\frac{\pi i}{2p}-1)$. – tywebb Jun 10 '23 at 12:56
  • I use \tiny in the last one because I DON'T want to make it larger. Larger would be lovely if everyone were viewing this on a 5 metre wide screen. But they aren't. – tywebb Jun 10 '23 at 13:10

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