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 ))


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

$p r o v e t h a t$ $\underset{k = 0}{\sum^{\infty}} \frac{{(k + 2)}^{2} x^{k}}{(k + 3)!} = \frac{e^{x}}{x^{3}} (x^{2} - x + 1) - \frac{x^{2} + 2}{2 x^{3}}$
	Answered by mind is power last updated on 12/Apr/20

	${(k + 2)}^{2} = (k + 3) (k + 2) - (k + 3) + 1$ $Σ \frac{x^{k}}{(k + 1)!} - Σ \frac{x^{k}}{(k + 2)!} + Σ \frac{x^{k}}{(k + 3)!}$ $\frac{1}{x} \sum_{k ⩾ 0} \frac{x^{k + 1}}{(k + 1)!} - \frac{1}{x^{2}} Σ \frac{x^{k + 2}}{(k + 2)!} + \frac{1}{x^{3}} Σ \frac{x^{k + 3}}{(k + 3)!}$ $= \frac{1}{x} (e^{x} - 1) - \frac{1}{x^{2}} (e^{x} - 1 - x) + \frac{1}{x^{3}} (e^{x} - 1 - x - \frac{x^{2}}{2})$ $= e^{x} (\frac{1}{x} - \frac{1}{x^{2}} + \frac{1}{x^{3}}) - \frac{1}{x^{3}} - \frac{1}{2 x^{}}$ $= \frac{e^{x}}{x^{3}} (1 - x + x^{2}) - \frac{2 + x^{2}}{2 x^{3}}$
	Commented by M±th+et£s last updated on 12/Apr/20

	$t h a n k y o u s i r .$ $y o u c a n t a k e a l o o k o n Q 88197$
	Commented by mind is power last updated on 12/Apr/20

	$y e a h w i t h e p l e a s u r$
	Commented by I want to learn more last updated on 12/Apr/20

	$Sir, please help with Q 88731$
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