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y. @waidotto

mathtod.online/@sr_ambivalence

$q:=1-p$とおくと
\begin{align*}
p\sum_{n=0}^\infty nq^{n-1}&=p\sum_{n=0}^\infty \frac{d}{dq}q^n\\&=p\frac{d}{dq}\sum_{n=0}^\infty q^n\\&=p\frac{d}{dq}\frac1{1-q}\\&=\frac{p}{(1-q)^2}=\frac{p}{p^2}=\frac1{p}
\end{align*}
ですね.

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