Question Number 50856 by peter frank last updated on 21/Dec/18
$${show}\:{that} \\ $$$$\underset{\mathrm{x}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}{xp}\left({x}\right)={np}\:{given}\: \\ $$$${that}\:{p}\left({x}\right)=^{\mathrm{n}} {C}_{\mathrm{x}} {p}^{{x}} {q}^{{n}−{x}} \\ $$
Answered by Smail last updated on 21/Dec/18
$${Q}\left({p}\right)=\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}{p}\left({x}\right)=\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{x}} {p}^{{x}} {q}^{{n}−{x}} =\left({p}+{q}\right)^{{n}} \\ $$$$\frac{{dQ}\left({p}\right)}{{dp}}=\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{x}} \left({xp}^{{x}−\mathrm{1}} \right){q}^{{n}−{x}} ={n}\left({p}+{q}\right)^{{n}−\mathrm{1}} \\ $$$$=\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}{x}\:^{{n}} {C}_{{x}} {p}^{{x}} {q}^{{n}−{x}} ×\frac{\mathrm{1}}{{p}}={n}\left({p}+{q}\right)^{{n}−\mathrm{1}} \\ $$$$=\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}{x}\left(^{{n}} {C}_{{x}} {p}^{{x}} {q}^{{n}−{x}} \right)={n}\left({p}+{q}\right)^{{n}−\mathrm{1}} {p} \\ $$$$\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}{xp}\left({x}\right)={np}\left({p}+{q}\right)^{{n}−\mathrm{1}} \\ $$
Commented by Smail last updated on 21/Dec/18
$${I}\:{think}\:{you}\:{are}\:{missing}\:{something}. \\ $$
Commented by peter frank last updated on 21/Dec/18
$${thank}\:{you} \\ $$
Commented by Smail last updated on 22/Dec/18
$${You}\:{are}\:{welcome} \\ $$