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Question Number 98096 by Algoritm last updated on 11/Jun/20

Answered by mathmax by abdo last updated on 11/Jun/20

S =Σ_(n=1) ^∞ ((3n−1)/2^(n−1) ) =_(n−1=p) Σ_(p=0) ^∞ ((3p+2)/2^p ) so S =3Σ_(n=0) ^∞ (n/2^n ) +2 Σ_(n=0) ^∞ (1/2^n ) we have Σ_(n=0) ^∞ (1/2^n ) =(1/(1−(1/2))) =2 let f(x) =Σ_(n=0) ^∞ nx^n with ∣x∣<1 we have Σ_(n=0) ^∞ x^n =(1/(1−x)) ⇒ Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒Σ_(n=1) ^∞ nx^n =(x/((1−x)^2 )) ⇒ Σ_(n=0) ^∞ n((1/2))^n =(1/(2(1−(1/2))^2 )) =(1/(2×(1/4))) =2 ⇒ S =3×2) +(2×2 S =10

$$\mathrm{S}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{3n}−\mathrm{1}}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\:=_{\mathrm{n}−\mathrm{1}=\mathrm{p}} \:\:\sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3p}+\mathrm{2}}{\mathrm{2}^{\mathrm{p}} } \\ $$$$\mathrm{so}\:\mathrm{S}\:=\mathrm{3}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{n}}{\mathrm{2}^{\mathrm{n}} }\:+\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:\:\:\mathrm{we}\:\mathrm{have}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\:=\mathrm{2} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{nx}^{\mathrm{n}} \:\:\:\:\:\:\mathrm{with}\:\mid\mathrm{x}\mid<\mathrm{1}\:\mathrm{we}\:\mathrm{have}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{nx}^{\mathrm{n}} \:=\frac{\mathrm{x}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\left.\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{n}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}} \:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}}\:=\mathrm{2}\:\Rightarrow\:\mathrm{S}\:=\mathrm{3}×\mathrm{2}\right)\:+\left(\mathrm{2}×\mathrm{2}\right. \\ $$$$\mathrm{S}\:=\mathrm{10} \\ $$

Commented by Algoritm last updated on 11/Jun/20

an elementary solution was needed 2S−S=?

$$\mathrm{an}\:\mathrm{elementary}\:\:\mathrm{solution}\:\mathrm{was}\:\mathrm{needed} \\ $$$$\mathrm{2S}−\mathrm{S}=? \\ $$

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