Advertisements
Advertisements
Question
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
Options
10
100
150
none of these
Advertisements
Solution
100
\[\lim_{x \to 1} \frac{x + x^2 + x^3 . . . x^n - n}{x - 1} = 5050\]
\[ \Rightarrow \lim_{x \to 1} \frac{\left( x - 1 \right)}{x - 1} + \frac{\left( x^2 - 1 \right)}{x - 1} + \frac{\left( x^3 - 1 \right)}{x - 1} . . . \frac{x^n - 1}{x - 2} = 5050\]
\[ \Rightarrow 1 + 2 + 3 . . . n = 5050 \left[ \because \frac{x^n - a^n}{x - a} = n a^{n - 1} \right]\]
\[ \Rightarrow \frac{n\left( n + 1 \right)}{2} = 5050\]
\[ \Rightarrow n\left( n + 1 \right) = 10100\]
\[ \Rightarrow n\left( n + 1 \right) = 100 \times 101\]
\[\text{ On comparing }: \]
\[n = 100\]
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]
\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{2}{x^2 - 2x} \right)\]
\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\]
\[\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{3}{x^2 - 3x} \right)\]
\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\]
\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\]
\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\]
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
\[\lim_{x \to 0} \left[ \frac{x^2}{\sin x^2} \right]\]
\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\]
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\]
\[\lim_{x \to 0} \frac{3 \sin^2 x - 2 \sin x^2}{3 x^2}\]
\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\]
\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]
\[\lim_{x \to 0} \frac{\sin 3x + 7x}{4x + \sin 2x}\]
\[\lim_{x \to 0} \frac{3 \sin x - \sin 3x}{x^3}\]
Evaluate the following limit:
\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]
\[\lim_{n \to \infty} \frac{\sin \left( \frac{a}{2^n} \right)}{\sin \left( \frac{b}{2^n} \right)}\]
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\] then \[\lim_{x \to 0} f\left( x \right) =\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] is equal at
\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to
\[\lim_{x \to \infty} \frac{\sqrt{x^2 - 1}}{2x + 1}\]
\[\lim_\theta \to \pi/2 \frac{1 - \sin \theta}{\left( \pi/2 - \theta \right) \cos \theta}\] is equal to
Evaluate the following limit:
`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.
Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
If f(x) = `{{:(1 if x "is rational"),(-1 if x "is rational"):}` is continuous on ______.
Evaluate the following limit :
`lim_(x->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`
Evaluate the following limit:
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
Evaluate the following limit.
`lim_(x->3)[sqrt(x + 6)/x]`
Evaluate the following limit.
`lim_(x->5)[(x^3 -125)/(x^5 - 3125)]`
Evaluate the Following limit:
`lim_(x->7)[((root(3)(x)-root(3)(7))(root(3)(x)+root(3)(7)))/(x-7)]`
Evaluate the following limit:
`\underset{x->3}{lim}[sqrt(x +6)/(x)]`
