Problem

5 pts
Suppose that $\sum_{n=0}^{\infty} c_{n}(x-2)^{n}$ is a power series with a finite. radius of convergence $R> 0$ such that the series converges for $x=5$ and diverges for $x=-5$. Which of the following are necessarily. true?
P. $R=5$
Q. $\sum_{n=0}^{\infty}(-1)^{n} c_{n}$ converges
$\sum_{n=0}^{\infty} 7^{n} c_{n}$ diverges
R only
$P, Q$, and $R$
None of them
Q only
$P$ and $Q$ only
Ponly
$P$ and $R$ only
$\mathrm{Q}$ and $\mathrm{R}$ only

Answer

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Answer

Final Answer: \(\boxed{\text{R only}}\)

Steps

Step 1 :The power series is centered at \(x=2\) and it converges for \(x=5\) and diverges for \(x=-5\). This means that the radius of convergence \(R\) is the distance from the center of the series to the point of convergence, which is \(5-2=3\). So, \(R=3\), not \(5\). Therefore, statement P is false.

Step 2 :For statement Q, we need to determine if the series \(\sum_{n=0}^{\infty}(-1)^{n} c_{n}\) converges. This is a series with alternating signs, and we can't determine its convergence without knowing the values of \(c_n\). Therefore, we can't say that statement Q is necessarily true.

Step 3 :For statement R, we need to determine if the series \(\sum_{n=0}^{\infty} 7^{n} c_{n}\) diverges. This is a power series with a common ratio of \(7\), which is greater than \(1\). According to the ratio test, a power series converges if the absolute value of the common ratio is less than \(1\) and diverges if it is greater than \(1\). Therefore, statement R is true.

Step 4 :So, the final answer is R only.

Step 5 :Final Answer: \(\boxed{\text{R only}}\)

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