Question

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

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Answer 1

Answer:

$\mathrm{\dfrac{1-n^2}{n!}\ or\ -\dfrac{n^2-1}{n!}}$

Step-by-step explanation:

Given :

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

To find :

The value of the given expression

Solution :

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

Multiplying n to denominator and numerator to second term and n(n-1) to third term, we get,

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

We know that,

$\boxed{\mathrm{x! = x.(x-1).(x-2)...3.2.1}}$

Thus,

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

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

$\implies \mathrm{\dfrac{1-\not{n}-n^2+\not{n}}{n!}}$

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

$\therefore \Bigg| \overline{\underline{\boxed{\mathbf{\dfrac{1}{n!} - \dfrac{1}{(n-1)!} - \dfrac{1}{(n-2)!} = \dfrac{1-n^2}{n!}}}}}\Bigg|$

Hope it helps!!