Math, asked by Anonymous, 6 hours ago

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

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Answered by IamIronMan0

Answer:

$\dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} \ - \ \dfrac{1}{(n - 2)!} \\ \\ = \dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} . \frac{n}{n} \ - \ \dfrac{1}{(n - 2)!} .\frac{n(n - 1)}{n(n - 1)} \\ \\ = \dfrac{1}{n!} \ - \ \dfrac{n}{n !} \ - \ \dfrac{n(n - 1)}{n !} \\ \\ = \dfrac{1}{n!} (1 - n - {n}^{2} + n) \\ \\ = \red{\dfrac{1 - {n}^{2} }{n!} }$

Answered by MichWorldCutiestGirl

SoLuTiOn,

$\begin{gathered}\dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} \ - \ \dfrac{1}{(n - 2)!} \\ \\ = \dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} . \frac{n}{n} \ - \ \dfrac{1}{(n - 2)!} .\frac{n(n - 1)}{n(n - 1)} \\ \\ = \dfrac{1}{n!} \ - \ \dfrac{n}{n !} \ - \ \dfrac{n(n - 1)}{n !} \\ \\ = \dfrac{1}{n!} (1 - n - {n}^{2} + n) \\ \\ = \pink{\dfrac{1 - {n}^{2} }{n!} }\end{gathered}$

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SoLuTiOn,

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