\(\frac{2\sqrt{2 }}{9801}{\displaystyle \sum _{k=0}^{\infty}}\frac{(4k)!(1103+26930k)}{(k!)^4{396}^{4k}}=\frac{1}{\pi}\)
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
+678 
