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@ganeshie8
sum of pdf is always 1, right?
\[1 = \sum_{x=0}^\infty k\left(\dfrac{4}{5}\right)^x = k\sum_{x=0}^\infty \left(\dfrac{4}{5}\right)^x\] So \(\displaystyle \dfrac{1}{\displaystyle \sum_{x=0}^\infty \left(\dfrac{4}{5}\right)^x} = k\). you can solve sum using the fact that \(\displaystyle \sum_{x=0}^\infty r^x = \dfrac{1}{1-r}\) (if i remember correctly lol)
does that help? @junyang96
oh I see. I totally forgot the sum to infinity formula. Thanks
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