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 _______.
Answer (integer)
40
Solution
$$\mathop {\lim }\limits_{x \to 1} {{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}$$ = 820
<br><br>As it is $\left( {{0 \over 0}} \right)$ form, Apply L'Hospital's Rule.
<br><br>$$\mathop {\lim }\limits_{x \to 1} \left( {{{1 + 2x + 3{x^2} + ... + n{x^{n - 1}}} \over 1}} \right)$$ = 820
<br><br>$\Rightarrow$ 1 + 2 + 3 + .....+ n = 820
<br><br>$\Rightarrow$ ${{n\left( {n + 1} \right)} \over 2}$ = 820
<br><br>$\Rightarrow$ n<sup>2</sup> + n – 1640 = 0
<br><br>$\Rightarrow$ (n – 40)(n + 41) = 0
<br><br>Since n $\in$ N, so n = 40.
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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