If _ x → 1 x + x ^ 2 + x ^ 3 + . . . + x ^ n - n x - 1 = 820, (n ∈...

\lim_{x \to 1}\dfrac{x + {x}^{2} + {x}^{3} + \ldots . + {x}^{n} - n}{x - 1} = 820$\left(\dfrac{0}{0}\right)$form

Applying L'Hospital rule

$\lim_{x \to 1}\dfrac{1 + 2x + 3{x}^{2} + \ldots . . + n{x}^{n - 1}}{1} = 820$

$⇒ 1 + 2 + 3 + \ldots \ldots + n = 820$

$⇒ \dfrac{n ( n + 1 )}{2} = 820$

$⇒ {n}^{2} + n = 1640$

$⇒ {n}^{2} + n - 1640 = 0$

$⇒ n = 40, - 41$

Since $n$ is natural number $⇒ n = 40$.