Question and Answers Forum

All Questions      Topic List

Probability and Statistics Questions

Previous in All Question      Next in All Question      

Previous in Probability and Statistics      Next in Probability and Statistics      

Question Number 105487 by Ar Brandon last updated on 29/Jul/20

Σ_(k=1) ^∞ k^2 (0.8)^(k−1)

$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{k}^{\mathrm{2}} \left(\mathrm{0}.\mathrm{8}\right)^{\mathrm{k}−\mathrm{1}} \\ $$

Answered by mathmax by abdo last updated on 29/Jul/20

let find S(x) =Σ_(n=1) ^∞ n^2 x^(n−1) 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=1) ^∞ nx^(n−1) =(d/dx){(x/((x−1)^2 ))} =(((x−1)^2 −2(x−1)x)/((x−1)^4 )) =((x−1−2x)/((x−1)^3 )) =((−x−1)/((x−1)^3 )) =((x+1)/((1−x)^3 )) ⇒S(x) =((x+1)/((1−x)^3 )) ⇒ Σ_(n=1) ^∞ n^2 (0,8)^(n−1) =Σ_(n=1) ^∞ n^2 ((4/5))^(n−1) =S((4/5)) =(((4/5)+1)/((1−(4/5))^3 )) =(9/5)×5^3 =25×9 =225

$$\mathrm{let}\:\mathrm{find}\:\mathrm{S}\left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{n}^{\mathrm{2}} \:\mathrm{x}^{\mathrm{n}−\mathrm{1}} \:\:\:\mathrm{with}\:\mid\mathrm{x}\mid<\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{nx}^{\mathrm{n}} \:=\frac{\mathrm{x}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \:=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\frac{\mathrm{x}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)\mathrm{x}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{4}} }\:=\frac{\mathrm{x}−\mathrm{1}−\mathrm{2x}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{−\mathrm{x}−\mathrm{1}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{x}+\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\mathrm{S}\left(\mathrm{x}\right)\:=\frac{\mathrm{x}+\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{n}^{\mathrm{2}} \left(\mathrm{0},\mathrm{8}\right)^{\mathrm{n}−\mathrm{1}} \:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{n}^{\mathrm{2}} \left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{n}−\mathrm{1}} \:=\mathrm{S}\left(\frac{\mathrm{4}}{\mathrm{5}}\right)\:=\frac{\frac{\mathrm{4}}{\mathrm{5}}+\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{9}}{\mathrm{5}}×\mathrm{5}^{\mathrm{3}} \:=\mathrm{25}×\mathrm{9}\:=\mathrm{225} \\ $$

Commented by Ar Brandon last updated on 29/Jul/20

Thanks

Commented by mathmax by abdo last updated on 29/Jul/20

you are welcome.

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com