Answer
The series converges for all real numbers \( x \). When \( x = \frac{5}{2} \), the sum to infinity of the series does not exist because it diverges. If the sum to infinity is \( \frac{49}{2} \), then \( x \) is approximately \( -1.183 \) or \( 4.183 \).
Solution
To analyze the series \( \sum x^{2}(x-3)^{2} \), we will address each part of the problem step by step.
### Part (1): Convergence of the Series
The series can be expressed as:
\[
\sum x^{2}(x-3)^{2} = \sum (x^{2}(x^{2} - 6x + 9)) = \sum (x^{4} - 6x^{3} + 9x^{2})
\]
This is a polynomial series, and polynomial series converge for all real values of \( x \). Therefore, the series converges for all \( x \in \mathbb{R} \).
### Part (2): Sum to Infinity for \( x = \frac{5}{2} \)
Substituting \( x = \frac{5}{2} \) into the series:
\[
\sum \left( \left( \frac{5}{2} \right)^{2} \left( \frac{5}{2} - 3 \right)^{2} \right)
\]
Calculating \( \left( \frac{5}{2} \right)^{2} \):
\[
\left( \frac{5}{2} \right)^{2} = \frac{25}{4}
\]
Calculating \( \left( \frac{5}{2} - 3 \right)^{2} \):
\[
\frac{5}{2} - 3 = \frac{5}{2} - \frac{6}{2} = -\frac{1}{2}
\]
\[
\left( -\frac{1}{2} \right)^{2} = \frac{1}{4}
\]
Now, substituting these values back into the series:
\[
\sum \left( \frac{25}{4} \cdot \frac{1}{4} \right) = \sum \frac{25}{16}
\]
Since this is a constant term, the sum diverges to infinity. Therefore, the sum to infinity of the series does not exist in the traditional sense, as it diverges.
### Part (3): Finding \( x \) for a Given Sum to Infinity
If the sum to infinity of the series is given as \( \frac{49}{2} \), we set up the equation:
\[
x^{2}(x-3)^{2} = \frac{49}{2}
\]
Expanding the left side:
\[
x^{2}(x^{2} - 6x + 9) = \frac{49}{2}
\]
\[
x^{4} - 6x^{3} + 9x^{2} = \frac{49}{2}
\]
Multiplying through by 2 to eliminate the fraction:
\[
2x^{4} - 12x^{3} + 18x^{2} - 49 = 0
\]
Now, we will solve this polynomial equation for \( x \).
Let's use the tool to solve the equation \( 2x^{4} - 12x^{3} + 18x^{2} - 49 = 0 \).
I'll proceed to solve this equation.
Solve the equation(The complex numbers system) by following steps:
- step0: Solve for \(x\) in the complex numbers system:
\(2x^{4}-12x^{3}+18x^{2}-49=0\)
- step1: Calculate:
\(\begin{align}&x\approx 4.183235\\&x\approx \frac{3}{2}-1.643091i\\&x\approx \frac{3}{2}+1.643091i\\&x\approx -1.183235\end{align}\)
- step2: Rewrite:
\(x_{1}\approx -1.183235,x_{2}\approx \frac{3}{2}+1.643091i,x_{3}\approx \frac{3}{2}-1.643091i,x_{4}\approx 4.183235\)
- step3: Remove the complex number(s):
\(x_{1}\approx -1.183235,x_{2}\approx 4.183235\)
The solutions to the equation \( 2x^{4} - 12x^{3} + 18x^{2} - 49 = 0 \) are approximately:
1. \( x_{1} \approx -1.183235 \)
2. \( x_{2} \approx 4.183235 \)
### Summary of Results:
1. The series converges for all \( x \in \mathbb{R} \).
2. The sum to infinity of the series does not exist as it diverges.
3. The values of \( x \) for which the sum to infinity equals \( \frac{49}{2} \) are approximately \( -1.183235 \) and \( 4.183235 \).
Reviewed and approved by the UpStudy tutoring team
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