21\sum_{i=1}^{p} i \frac{\binom{p}{i}}{n^p}\left ( \frac{n-1}{p} \right )^{p-i} = \frac{p}{n}
Prove that \sum_{i=1}^{p}i \binom{p}{i}(n-1)^{p-i} = p n^{p-1}
n is real, p is natural number
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1\begin{align*} & i{ p \choose i} = p{ p-1 \choose i-1}\\ & \implies \textup{L.H.S is }, \\ & p\Sigma { p-1 \choose i-1} (n-1)^{p-i} \\ & p ((n-1)+1)^{p-1} \\ & \boxed{p(n)^{p-1}} \\ &= R.H.S \\ & \\ & \\ & \\ & \end{align*}