QJEEM20Ma48ac: JEE Main 2020 Mathematics - Class 12 JEEM20Ma48ac Solution: If = 820, (n N) then the value of n is equal to _______.

$\mathop {\lim }\limits_{x \to 1} {{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}$ = 820

As it is $\left( {{0 \over 0}} \right)$ form, Apply L'Hospital's Rule.

$\mathop {\lim }\limits_{x \to 1} \left( {{{1 + 2x + 3{x^2} + ... + n{x^{n - 1}}} \over 1}} \right)$ = 820

$\Rightarrow$ 1 + 2 + 3 + .....+ n = 820

$\Rightarrow$ ${{n\left( {n + 1} \right)} \over 2}$ = 820

$\Rightarrow$ n² + n – 1640 = 0

$\Rightarrow$ (n – 40)(n + 41) = 0

Since n $\in$ N, so n = 40.