Given that f(r)=(r+1)! ∙ r, show that f(r)−f(r−1)=r!(r^2 +1). Hence or otherwise, show that 2! ∙ 5+3! ∙ 10+4! ∙ 17+......n!(n^2 +1)=(n+1)! ∙ (n−2)


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Question Number 143393 by ZiYangLee last updated on 13/Jun/21

$Given that f (r) = (r + 1)! \cdot r, show that$ $f (r) - f (r - 1) = r! (r^{2} + 1) .$ $Hence or otherwise, show that$ $2! \cdot 5 + 3! \cdot 10 + 4! \cdot 17 + . . . . . . n! (n^{2} + 1) = (n + 1)! \cdot (n - 2)$
	Answered by TheHoneyCat last updated on 13/Jun/21

	$f (r) - f (r - 1)$ $= (r + 1)! \cdot r - r! \cdot (r - 1)$ $= (r + 1) \cdot r! \cdot r - (r - 1) \cdot r!$ $= r! \cdot (r^{2} + r - r + 1)$ $= r! \cdot (r^{2} + 1)$
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