Again we have

$\begin{array}{ll}{\mu \prime}_{3}\hfill & =\sum _{x=0}^{n}{x}^{3}p(x)\hfill \\ =\sum _{x=0}^{n}[x(x-1)(x-2)+3x(x-1)+x]p(x)\hfill \\ (\mathrm{since}{x}^{3}=x(x-1)(x\u20132)+3x(x\u20131)+x)\hfill \\ =\sum _{x=0}^{n}[x(x-1)(x-2)+3\sum _{x=0}^{n}x(x-1)+p(x)+\sum _{x=0}^{n}xP(x)\hfill \\ =n(n\u20131)(n\u20132){p}^{3}{(q+p)}^{n\u20133}+3n(n\u20131){p}^{2}{(q+p)}^{n\u20132}+np{(q+p)}^{n\u20131}\hfill \\ =n(n\u20131)(n\u20132){p}^{3}+3n(n\u20131){p}^{2}+np\hfill \end{array}$

Since ${\mu}_{3}={\mu \prime}_{3}-3{\mu \prime}_{2}{\mu \prime}_{1}+2{\mu}_{1}^{\prime 3}$

We get

$\begin{array}{ll}{\mu}_{3}\hfill & =n(n\u20131)(n\u20132){p}^{3}+3n(n\u20131){p}^{2}+np\u20133({n}^{2}{p}^{2}\u2013n{p}^{2}+np)(np)+2{n}^{3}{p}^{3}\hfill \\ =npq(q\u2013p)\hfill \end{array}$

Similarly we get

${\mu \prime}_{4}=n(n-1)(n-2)$

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