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Question: To solve;

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$\dashrightarrow \ \sf \dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} \ - \ \dfrac{1}{(n - 2)!}$

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Answer 1

Step-by-step explanation:

$\dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} \ - \ \dfrac{1}{(n - 2)!}$

${ \implies\dfrac{1}{n(n - 1)(n - 2)!} \ - \ \dfrac{1}{(n - 1)(n - 2)!} \ - \ \dfrac{1}{(n - 2)!}}$

${\implies \dfrac{1}{(n - 2)!}\left[ \dfrac{1}{n(n - 1)} - \dfrac{1}{(n - 1)} - \dfrac{1}{1} \right]}$

${\implies \dfrac{1}{(n - 2)!}\left[ \dfrac{1 - n - n(n - 1)}{n(n - 1)} \right]}$

${\implies \dfrac{1}{(n - 2)!}\left[ \dfrac{1 - n - {n}^{2} + n}{n(n - 1)} \right]}$

${\implies \dfrac{1}{(n - 2)!}\left[ \dfrac{1-{n}^{2}}{n(n - 1)} \right]}$

${\implies\dfrac{1-{n}^{2}}{n(n - 1)(n - 2)!}}$

${\implies\dfrac{1-{n}^{2}}{n !}}$

Therefore,

$\boxed{ \dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} \ - \ \dfrac{1}{(n - 2)!} = \dfrac{1-{n}^{2}}{n !}}$

Note:

• $n! = n(n-1)!$

and similarly we can write:

• $n! = n (n-1) (n-2)!$
• $n! = n (n-1) (n-2) (n-3)!$

... and so on