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Question Number 73231 by mathmax by abdo last updated on 08/Nov/19

find the sum of Σ_(n=0) ^∞ (n^2 −3n+1)e^(−n)

$${find}\:{the}\:{sum}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{1}\right){e}^{−{n}} \\ $$

Commented by mathmax by abdo last updated on 09/Nov/19

we have Σ_(n=0) ^∞ (n^2 −3n+1)e^(−n) =Σ_(n=0) ^∞ n^2 (e^(−1) )^n −3Σ_(n=0) ^∞ n(e^(−1) )^n +Σ_(n=0) ^∞ e^(−n) first Σ_(n=0) ^∞ e^(−n) =Σ_(n=0) ^∞ (e^(−1) )^n =(1/(1−e^(−1) )) =(e/(e−1)) let calculate f(x)=Σ_(n=0) ^∞ nx^n and g(x)=Σ_(n=0) ^∞ n^2 x^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) ^∞ n x^n =(x/((1−x)^2 )) ⇒Σ_(n=0) ^∞ ne^(−n) =(e^(−1) /((1−e^(−1) )^2 )) =(e^2 /((e−1)^2 )) also Σ_(n=1) ^∞ n^2 x^(n−1) =(((1−x)^2 −x(−2)(1−x))/((1−x)^4 )) =((1−x+2x)/((1−x)^3 )) =((x+1)/((1−x)^3 )) ⇒ Σ_(n=0) ^∞ n^2 x^n =((x^2 +x)/((1−x)^3 )) ⇒Σ_(n=0) ^∞ n^2 e^(−n) =((e^(−2) +e^(−1) )/((1−e^(−1) )^3 )) =((e+e^2 )/((e−1)^3 )) ⇒Σ_(n=0) ^∞ (n^2 −3n +1)e^(−n) =((e+e^2 )/((e−1)^3 ))−((3e^2 )/((e−1)^2 )) +(e/(e−1)) =((e+e^2 −3e^2 (e−1)+e(e−1)^2 )/((e−1)^3 )) =((e+e^2 −3e^3 +3e^2 +e(e^2 −2e +1))/((e−1)^3 )) =((−3e^3 +4e^2 +e+e^3 −2e^2 +e)/((e−1)^3 )) =((−2e^3 +2e^2 +2e)/((e−1)^3 ))

$${we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{1}\right){e}^{−{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} {n}^{\mathrm{2}} \left({e}^{−\mathrm{1}} \right)^{{n}} −\mathrm{3}\sum_{{n}=\mathrm{0}} ^{\infty} \:{n}\left({e}^{−\mathrm{1}} \right)^{{n}} \\ $$$$+\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}} \:\:\:{first}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}} =\sum_{{n}=\mathrm{0}} ^{\infty} \left({e}^{−\mathrm{1}} \right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }\:=\frac{{e}}{{e}−\mathrm{1}} \\ $$$${let}\:{calculate}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{nx}^{{n}} \:\:{and}\:{g}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} {n}^{\mathrm{2}} {x}^{{n}} \:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\:{x}^{{n}} \:=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:{ne}^{−{n}} \:=\frac{{e}^{−\mathrm{1}} }{\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)^{\mathrm{2}} }\:=\frac{{e}^{\mathrm{2}} }{\left({e}−\mathrm{1}\right)^{\mathrm{2}} }\:{also} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} \:{x}^{{n}−\mathrm{1}} \:=\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −{x}\left(−\mathrm{2}\right)\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\frac{\mathrm{1}−{x}+\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:=\frac{{x}+\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{n}^{\mathrm{2}} \:{x}^{{n}} \:=\frac{{x}^{\mathrm{2}} \:+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:{n}^{\mathrm{2}} {e}^{−{n}} \:=\frac{{e}^{−\mathrm{2}} \:+{e}^{−\mathrm{1}} }{\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)^{\mathrm{3}} } \\ $$$$=\frac{{e}+{e}^{\mathrm{2}} }{\left({e}−\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}^{\mathrm{2}} −\mathrm{3}{n}\:+\mathrm{1}\right){e}^{−{n}} \:=\frac{{e}+{e}^{\mathrm{2}} }{\left({e}−\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{3}{e}^{\mathrm{2}} }{\left({e}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{e}}{{e}−\mathrm{1}} \\ $$$$=\frac{{e}+{e}^{\mathrm{2}} −\mathrm{3}{e}^{\mathrm{2}} \left({e}−\mathrm{1}\right)+{e}\left({e}−\mathrm{1}\right)^{\mathrm{2}} }{\left({e}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{{e}+{e}^{\mathrm{2}} −\mathrm{3}{e}^{\mathrm{3}} +\mathrm{3}{e}^{\mathrm{2}} +{e}\left({e}^{\mathrm{2}} −\mathrm{2}{e}\:+\mathrm{1}\right)}{\left({e}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{−\mathrm{3}{e}^{\mathrm{3}} \:+\mathrm{4}{e}^{\mathrm{2}} \:+{e}+{e}^{\mathrm{3}} −\mathrm{2}{e}^{\mathrm{2}} \:+{e}}{\left({e}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}{e}^{\mathrm{3}} +\mathrm{2}{e}^{\mathrm{2}} +\mathrm{2}{e}}{\left({e}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Answered by mind is power last updated on 08/Nov/19

n^2 −3n+1=(n+1)(n+2)−6(n+1)+5 Σ_(n≥0) (n+1)(n+2)e^(−n) −6Σ_(n≥0) (n+1)e^(−n) +5Σ_(n≥0) e^(−n) (1/(1−x))=Σx^k ⇒(1/((1−x)^2 )).Σ_(k≥1) kx^(k−1) =Σ_(k≥0) (k+1)x^k x=e^(−1) ⇒Σe^(−n) =(1/(1−e^(−1) ))=(e/(e−1)) ⇒x=e^(−1) ⇒(1/((1−e^(−1) )^2 ))=Σ_(n≥0) (n+1)e^(−n) =(e^2 /((e−1))) (1/((1−x)^2 ))=Σ_(k≥0) (k+1)x^k ⇒Σ_(k≥1) k(k+1)x^(k−1) =(2/((1−x)^3 ))=Σ_(k≥0) (k+1)(k+2)x^k x=e^(−1) ⇒Σ_(n≥0) (n+1)(n+2)e^(−n) =(2/((1−e^(−1) )^3 ))=((2e^3 )/((e−1)^3 )) so we get ((2e^3 )/((e−1)^3 ))−((6e^2 )/((e−1)^2 ))+((5e)/((e−1))) =((e^3 −4e^2 +5e)/((e−1)^3 ))

$$\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{1}=\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)−\mathrm{6}\left(\mathrm{n}+\mathrm{1}\right)+\mathrm{5} \\ $$$$\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\mathrm{e}^{−\mathrm{n}} −\mathrm{6}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{e}^{−\mathrm{n}} +\mathrm{5}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\mathrm{e}^{−\mathrm{n}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}=\Sigma\mathrm{x}^{\mathrm{k}} \Rightarrow\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }.\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\mathrm{kx}^{\mathrm{k}−\mathrm{1}} =\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\left(\mathrm{k}+\mathrm{1}\right)\mathrm{x}^{\mathrm{k}} \\ $$$$\mathrm{x}=\mathrm{e}^{−\mathrm{1}} \Rightarrow\Sigma\mathrm{e}^{−\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{1}−\mathrm{e}^{−\mathrm{1}} }=\frac{\mathrm{e}}{\mathrm{e}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{e}^{−\mathrm{1}} \Rightarrow\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{e}^{−\mathrm{1}} \right)^{\mathrm{2}} }=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{e}^{−\mathrm{n}} =\frac{\mathrm{e}^{\mathrm{2}} }{\left(\mathrm{e}−\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\left(\mathrm{k}+\mathrm{1}\right)\mathrm{x}^{\mathrm{k}} \Rightarrow\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)\mathrm{x}^{\mathrm{k}−\mathrm{1}} =\frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{3}} }=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{k}+\mathrm{2}\right)\mathrm{x}^{\mathrm{k}} \\ $$$$\mathrm{x}=\mathrm{e}^{−\mathrm{1}} \Rightarrow\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\mathrm{e}^{−\mathrm{n}} =\frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{e}^{−\mathrm{1}} \right)^{\mathrm{3}} }=\frac{\mathrm{2e}^{\mathrm{3}} }{\left(\mathrm{e}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{get}\:\frac{\mathrm{2e}^{\mathrm{3}} }{\left(\mathrm{e}−\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{6e}^{\mathrm{2}} }{\left(\mathrm{e}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{5e}}{\left(\mathrm{e}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{e}^{\mathrm{3}} −\mathrm{4e}^{\mathrm{2}} +\mathrm{5e}}{\left(\mathrm{e}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 08/Nov/19

thankx sir.

$${thankx}\:{sir}. \\ $$

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