The series $\sum_{m=0}^{\infty} \frac{1}{4^m}(x-1)^{2m}$ converges for

Question

Prepp Team · Accepted Answer

Series Convergence Analysis

The given series is $\sum_{m=0}^{\infty} \frac{1}{4^m}(x-1)^{2m}$. We need to find the interval for $x$ where this series converges.

The series can be rewritten by combining terms:

$ \sum_{m=0}^{\infty} \frac{1}{4^m}(x-1)^{2m} = \sum_{m=0}^{\infty} \left(\frac{(x-1)^2}{4}\right)^m $

This is a geometric series of the form $\sum_{m=0}^{\infty} r^m$, where the common ratio is:

$ r = \frac{(x-1)^2}{4} $

A geometric series converges if and only if the absolute value of its common ratio is less than 1:

$ |r| < 1 $ $ \left|\frac{(x-1)^2}{4}\right| < 1 $

Since $(x-1)^2$ is always non-negative, the absolute value is redundant:

$ \frac{(x-1)^2}{4} < 1 $

Multiply both sides by 4:

$ (x-1)^2 < 4 $

Take the square root of both sides:

$ \sqrt{(x-1)^2} < \sqrt{4} $ $ |x-1| < 2 $

This inequality means:

$ -2 < x-1 < 2 $

Add 1 to all parts of the inequality:

$ -2 + 1 < x-1 + 1 < 2 + 1 $ $ -1 < x < 3 $

The series converges for the interval $-1 < x < 3$. This corresponds to Option B.

Prepp Team · Answer

$-2

Prepp Team · Answer

$-3

Prepp Team · Answer

$x

List-1	List-2
P The sum of the series $\sum_{n=1}^\infty \frac{1}{(n+2)(n+1)}$ is equal to	I $\frac{3}{2}$
Q $\lim_{x \to 0} \left( \frac{3}{x^2} \int_0^x \sin(t) dt \right)$ is equal to	II $1$
R Let $\frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx)$ be the Fourier series expansion of the function $f(x) = \frac{1}{2} \sin x - \frac{1}{2} \cos x + \frac{1}{\sqrt{2}} \sin 2x, x \in [0, 2\pi]$. Then, $\sum_{n=0}^\infty (a_n^2 + b_n^2)$ is equal to	III $\frac{1}{2}$