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$\frac{2\sqrt{2 }}{9801}{\displaystyle \sum _{k=0}^{\infty}}\frac{(4k)!(1103+26930k)}{(k!)^4{396}^{4k}}=\frac{1}{\pi}$

Straight Oct 21, 2021

Guest · Answer 1 · 2021-11-03T01:24:32

can someone help me with this

There are 18 red markers and 10 yellow markers. The ratio of red to yellow.

what is the answer