The limit of sum_{n=1}^{1000} (-1)^{n} x^{n} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $S = \sum_{n=1}^{1000} (-1)^{n} x^{n} = -x + x^{2} - x^{3} + x^{4} - \dots + x^{1000}$.This is a finite geometric series with first term $a =

Vedclass · Accepted Answer

exists and approaches to $+\infty$

Vedclass · Answer

(C) Let $S = \sum_{n=1}^{1000} (-1)^{n} x^{n} = -x + x^{2} - x^{3} + x^{4} - \dots + x^{1000}$.This is a finite geometric series with first term $a = -x$,common ratio $r = -x$,and number of terms $n = 1000$.The sum is given by $S = a \frac{r^{n} - 1}{r - 1} = (-x) \frac{(-x)^{1000} - 1}{-x - 1} = (-x) \frac{x^{1000} - 1}{-(x + 1)} = x \frac{x^{1000} - 1}{x + 1} = \frac{x^{1001} - x}{x + 1}$.Now,we evaluate the limit as $x \rightarrow \infty$:$\lim_{x \rightarrow \infty} \frac{x^{1001} - x}{x + 1} = \lim_{x \rightarrow \infty} \frac{x^{1001}(1 - \frac{1}{x^{1000}})}{x(1 + \frac{1}{x})} = \lim_{x \rightarrow \infty} x^{1000} = +\infty$.

The limit of $\sum_{n=1}^{1000} (-1)^{n} x^{n}$ as $x \rightarrow \infty$ is:

Similar Questions

If $[x]$ is the greatest integer function,then $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)=$

$\lim _{x}$ ${\rightarrow 0^{+}} \frac{\tan \left(5(x)^{\frac{1}{3}}\right) \log _e\left(1+3 x^2\right)}{\left(\tan ^{-1} 3 \sqrt{x}\right)^2\left(e^{5(x)^{\frac{4}{3}}}-1\right)}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{|x|}{x} = $

If $l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^3 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \left( \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} \right)$,find the quadratic equation whose roots are $l$ and $m$.

The value of $\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^{2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module