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Question Number 33088 by Joel578 last updated on 10/Apr/18
Find the value of Σ_(n=1) ^∞ (n^2 /2^(n−1) )
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{2}} }{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$
Commented by abdo imad last updated on 10/Apr/18
let consider the serie s(x)= Σ_(n=0) ^∞ x^n with ∣x∣<1 we have s(x)=(1/(1−x)) ⇒ s^′ (x)= (1/((1−x)^2 )) ⇒Σ_(n=1) ^∞ n x^(n−1) =(1/((1−x)^2 )) also s^(′′) (x)=−((2(−1)(1−x))/((1−x)^4 )) =(2/((1−x)^3 )) ⇒ Σ_(n=2) ^∞ n(n−1)x^(n−2) = (2/((1−x)^3 )) ⇒ Σ_(n=2) ^∞ n(n−1)x^(n−1) =((2x)/((1−x)^3 )) ⇒ Σ_(n=2) ^∞ n^2 x^(n−1) =((2x)/((1−x)^3 )) +Σ_(n=2) ^∞ nx^(n−1) =((2x)/((1−x)^3 )) +(1/((1−x)^2 )) −1 ⇒Σ_(n=1) ^∞ n^2 x^(n−1) =((2x)/((1−x)^3 )) +(1/((1−x)^2 )) after we take x=(1/2) ⇒ Σ_(n=1) ^∞ (n^2 /2^(n−1) ) = (1/(((1/2))^3 )) + (1/(((1/2))^2 )) = 8 +4 =12 .
$${let}\:{consider}\:{the}\:{serie}\:{s}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:\:\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$${we}\:{have}\:{s}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow\:{s}^{'} \left({x}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} {n}\:{x}^{{n}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\:{also}\:{s}^{''} \left({x}\right)=−\frac{\mathrm{2}\left(−\mathrm{1}\right)\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\frac{\mathrm{2}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} {n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} \:=\:\frac{\mathrm{2}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} {n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{1}} \:\:=\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} \:=\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:+\sum_{{n}=\mathrm{2}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \\ $$$$=\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:−\mathrm{1}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} \:{x}^{{n}−\mathrm{1}} \:=\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${after}\:{we}\:{take}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{2}} }{\mathrm{2}^{{n}−\mathrm{1}} }\:\:=\:\:\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }\:\:+\:\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:=\:\mathrm{8}\:+\mathrm{4}\:=\mathrm{12}\:\:. \\ $$$$ \\ $$
Commented by abdo imad last updated on 10/Apr/18
remark ★we have 2 Σ_(n=1) ^∞ (n^2 /2^n ) =12 ⇒ Σ_(n=1) ^∞ (n^2 /2^n ) =6 .
$${remark}\:\:\bigstar{we}\:{have}\:\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{n}^{\mathrm{2}} }{\mathrm{2}^{{n}} }\:=\mathrm{12}\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{2}} }{\mathrm{2}^{{n}} }\:\:=\mathrm{6}\:. \\ $$
Commented by Joel578 last updated on 13/Apr/18
Thank you very much
$${Thank}\:{you}\:{very}\:{much} \\ $$

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