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Question Number 88723 by M±th+et£s last updated on 12/Apr/20

prove that Σ_(k=0) ^∞ (((k+2)^2 x^k )/((k+3)!))=(e^x /x^3 )(x^2 −x+1)−((x^2 +2)/(2x^3 ))

$${prove}\:{that} \\ $$$$ \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({k}+\mathrm{2}\right)^{\mathrm{2}} {x}^{{k}} }{\left({k}+\mathrm{3}\right)!}=\frac{{e}^{{x}} }{{x}^{\mathrm{3}} }\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)−\frac{{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}^{\mathrm{3}} } \\ $$

Answered by mind is power last updated on 12/Apr/20

(k+2)^2 =(k+3)(k+2)−(k+3)+1 Σ(x^k /((k+1)!))−Σ(x^k /((k+2)!))+Σ(x^k /((k+3)!)) (1/x)Σ_(k≥0) (x^(k+1) /((k+1)!))−(1/x^2 )Σ(x^(k+2) /((k+2)!))+(1/x^3 )Σ(x^(k+3) /((k+3)!)) =(1/x)(e^x −1)−(1/x^2 )(e^x −1−x)+(1/x^3 )(e^x −1−x−(x^2 /2)) =e^x ((1/x)−(1/x^2 )+(1/x^3 ))−(1/x^3 )−(1/(2x^ )) =(e^x /x^3 )(1−x+x^2 )−((2+x^2 )/(2x^3 ))

$$\left({k}+\mathrm{2}\right)^{\mathrm{2}} =\left({k}+\mathrm{3}\right)\left({k}+\mathrm{2}\right)−\left({k}+\mathrm{3}\right)+\mathrm{1} \\ $$$$\Sigma\frac{{x}^{{k}} }{\left({k}+\mathrm{1}\right)!}−\Sigma\frac{{x}^{{k}} }{\left({k}+\mathrm{2}\right)!}+\Sigma\frac{{x}^{{k}} }{\left({k}+\mathrm{3}\right)!} \\ $$$$\frac{\mathrm{1}}{{x}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{{k}+\mathrm{1}} }{\left({k}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\Sigma\frac{{x}^{{k}+\mathrm{2}} }{\left({k}+\mathrm{2}\right)!}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\Sigma\frac{{x}^{{k}+\mathrm{3}} }{\left({k}+\mathrm{3}\right)!} \\ $$$$=\frac{\mathrm{1}}{{x}}\left({e}^{{x}} −\mathrm{1}\right)−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left({e}^{{x}} −\mathrm{1}−{x}\right)+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\left({e}^{{x}} −\mathrm{1}−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$={e}^{{x}} \left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}{x}^{} } \\ $$$$=\frac{{e}^{{x}} }{{x}^{\mathrm{3}} }\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)−\frac{\mathrm{2}+{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{3}} } \\ $$$$ \\ $$

Commented by M±th+et£s last updated on 12/Apr/20

thank you sir . you can take a look on Q88197

$${thank}\:{you}\:{sir}\:. \\ $$$${you}\:{can}\:{take}\:{a}\:{look}\:{on}\:{Q}\mathrm{88197}\: \\ $$

Commented by mind is power last updated on 12/Apr/20

yeah withe pleasur

$${yeah}\:{withe}\:{pleasur}\: \\ $$

Commented by I want to learn more last updated on 12/Apr/20

Sir, please help with Q88731

$$\mathrm{Sir},\:\:\mathrm{please}\:\mathrm{help}\:\mathrm{with}\:\:\:\mathrm{Q88731} \\ $$

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