Question

Let (a_n) be a sequence of real numbers such that the series \(\sum\limits_{n=0}^{\infin}a_n(x-2)^n\) converges at x = −5. Then this series also converges at

• A. x = 9
• B. x = 12
• C. x = 5
• D. x = −6

Answer 1

The Correct Option is C

To solve this problem, we need to understand the convergence of the given power series:

\(\sum\limits_{n=0}^{\infin}a_n(x-2)^n\)

It is given that this series converges at \(x = -5\). We need to determine for which of the given options the series also converges.

For a power series of the form:

\(\sum\limits_{n=0}^{\infin} a_n (x-c)^n\) 

convergence depends on the distance from \(x\) to the center \(c\). In this question, \(c = 2\).

The series converges within a radius \(R\), where:

\(|x - c| < R\)

Since the series converges at \(x = -5\), the radius of convergence \(R\) must satisfy:

\(|-5 - 2| = 7 \leq R\)

This implies \(R \geq 7\).

Let's check each option to see if they fall within this radius from the center:

• Option 1: \(x = 9\)
• Option 2: \(x = 12\)
• Option 3: \(x = 5\)
• Option 4: \(x = -6\)

Thus, the given series also converges at \(x = 5\).