JEE MAIN - Mathematics (2020 - 2nd September Morning Slot - No. 9)
If $$\mathop {\lim }\limits_{x \to 1} {{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}$$ = 820,
(n $$ \in $$ N) then the value of n is equal to _______.
(n $$ \in $$ N) then the value of n is equal to _______.
Answer
40
Explanation
$$\mathop {\lim }\limits_{x \to 1} {{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}$$ = 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.
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.
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