If y = x - x^2 + x^3 - x^4 + dots infty,then | vedclass

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

Question

(D) Given the infinite geometric series: $y = x - x^2 + x^3 - x^4 + \dots \infty$ This is a geometric series with first term $a = x$ and common ratio

Vedclass · Accepted Answer

$\frac{y}{1 - y}$

Vedclass · Answer

(D) Given the infinite geometric series: $y = x - x^2 + x^3 - x^4 + \dots \infty$ This is a geometric series with first term $a = x$ and common ratio $r = -x$. The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$. Substituting the values: $y = \frac{x}{1 - (-x)} = \frac{x}{1 + x}$. Now,solve for $x$: $y(1 + x) = x$ $y + xy = x$ $y = x - xy$ $y = x(1 - y)$ $x = \frac{y}{1 - y}$

If $y = x - x^2 + x^3 - x^4 + \dots \infty$,then the value of $x$ is equal to:

Similar Questions

If the first term of a Geometric Progression $(GP)$ is $1$ and the sum of its third and fifth terms is $90$,find the common ratio.

Find the fractional value of $0.125125125 \dots$

What is the geometric mean of the observations $2, 4, 8, 16, 32, 64$?

The value of ${4^{1/3}} \cdot {4^{1/9}} \cdot {4^{1/27}} \cdots \infty$ is

$A$ geometric progression consists of positive terms. If each term is equal to the sum of the next two terms,what is the common ratio of the progression?

Vedclass Test Series

Exam Paper Generator

Online Exam Module