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Question Number 50856 by peter frank last updated on 21/Dec/18

show that Σ_(x=0) ^n xp(x)=np given that p(x)=^n C_x p^x q^(n−x)

$${show}\:{that} \\ $$$$\underset{\mathrm{x}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}{xp}\left({x}\right)={np}\:{given}\: \\ $$$${that}\:{p}\left({x}\right)=^{\mathrm{n}} {C}_{\mathrm{x}} {p}^{{x}} {q}^{{n}−{x}} \\ $$

Answered by Smail last updated on 21/Dec/18

Q(p)=Σ_(x=0) ^n p(x)=Σ_(x=0) ^n ^n C_x p^x q^(n−x) =(p+q)^n ((dQ(p))/dp)=Σ_(x=0) ^n ^n C_x (xp^(x−1) )q^(n−x) =n(p+q)^(n−1) =Σ_(x=0) ^n x^n C_x p^x q^(n−x) ×(1/p)=n(p+q)^(n−1) =Σ_(x=0) ^n x(^n C_x p^x q^(n−x) )=n(p+q)^(n−1) p Σ_(x=0) ^n xp(x)=np(p+q)^(n−1)

$${Q}\left({p}\right)=\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}{p}\left({x}\right)=\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{x}} {p}^{{x}} {q}^{{n}−{x}} =\left({p}+{q}\right)^{{n}} \\ $$$$\frac{{dQ}\left({p}\right)}{{dp}}=\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{x}} \left({xp}^{{x}−\mathrm{1}} \right){q}^{{n}−{x}} ={n}\left({p}+{q}\right)^{{n}−\mathrm{1}} \\ $$$$=\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}{x}\:^{{n}} {C}_{{x}} {p}^{{x}} {q}^{{n}−{x}} ×\frac{\mathrm{1}}{{p}}={n}\left({p}+{q}\right)^{{n}−\mathrm{1}} \\ $$$$=\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}{x}\left(^{{n}} {C}_{{x}} {p}^{{x}} {q}^{{n}−{x}} \right)={n}\left({p}+{q}\right)^{{n}−\mathrm{1}} {p} \\ $$$$\underset{{x}=\mathrm{0}} {\overset{{n}} {\sum}}{xp}\left({x}\right)={np}\left({p}+{q}\right)^{{n}−\mathrm{1}} \\ $$

Commented by Smail last updated on 21/Dec/18