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Part 1
Find the power series representation for g centered at 0 by differentiating or integrating the power series for f. Give the interval of convergence for the resulting series.
​g(x)equalsStartFraction 21 Over left parenthesis 1 minus 21 x right parenthesis squared EndFraction
​, ​f(x)equalsStartFraction 1 Over 1 minus 21 x EndFraction
Question content area bottom
Part 1
The power series representation for g is Summation from k equals 1 to infinity left parenthesis nothing right parenthesis
  
.

To find the power series representation for \( g(x) \), we start with the given functions:

1. \( f(x) = \frac{1}{1 - 21x} \)
2. \( g(x) = \frac{21}{(1 - 21x)^2} \)

### Step 1: Power Series for \( f(x) \)
The power series for \( f(x) \) is derived from the geometric series formula:

\[
\frac{1}{1 - r} = \sum_{k=0}^{\infty} r^k \quad \text{(for } |r| < 1 \text{)}.
\]

Substitute \( r = 21x \):
\[
f(x) = \frac{1}{1 - 21x} = \sum_{k=0}^{\infty} (21x)^k = \sum_{k=0}^{\infty} 21^k x^k, \quad \text{for } |21x| < 1 \text{ or } |x| < \frac{1}{21}.
\]

### Step 2: Derivative of \( f(x) \) to Obtain \( g(x) \)
The derivative of \( f(x) \) with respect to \( x \) is:
\[
f'(x) = \frac{d}{dx} \left( \frac{1}{1 - 21x} \right) = \frac{21}{(1 - 21x)^2}.
\]

Since \( g(x) = 21 \cdot f'(x) \), we have:
\[
g(x) = 21 \cdot \frac{21}{(1 - 21x)^2} = \frac{21}{(1 - 21x)^2}.
\]

### Step 3: Power Series for \( g(x) \)
To find the power series for \( g(x) \), differentiate the power series for \( f(x) \):
\[
f(x) = \sum_{k=0}^{\infty} 21^k x^k.
\]

Differentiating term by term:
\[
f'(x) = \sum_{k=1}^{\infty} k \cdot 21^k x^{k-1}.
\]

Multiply by 21 to account for \( g(x) = 21 \cdot f'(x) \):
\[
g(x) = 21 \cdot f'(x) = \sum_{k=1}^{\infty} k \cdot 21^{k+1} x^{k-1}.
\]

Rewrite the power series by shifting the index \( k \to k+1 \):
\[
g(x) = \sum_{k=1}^{\infty} k \cdot 21^{k+1} x^k.
\]

Thus, the power series representation for \( g(x) \) is:
\[
g(x) = \sum_{k=1}^{\infty} k \cdot 21^k x^k.
\]

### Step 4: Interval of Convergence
The power series for \( g(x) \) converges for the same \( |x| < \frac{1}{21} \), since it was derived from \( f(x) \) by differentiation.

### Final Answer
The power series representation for \( g(x) \) is:
\[
g(x) = \sum_{k=1}^{\infty} k \cdot 21^k x^k,
\]
and the interval of convergence is:
\[
|x| < \frac{1}{21}.
\]

Would you like additional explanations or derivations for any part? 

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#### 5 Related Questions:
1. How does differentiation affect the interval of convergence of a power series?
2. What is the general formula for the \( n \)-th derivative of a geometric series?
3. Can the power series for \( g(x) \) be integrated to recover \( f(x) \)?
4. What is the significance of the radius of convergence in applications of power series?
5. How do you find the nth term of a power series derived from a given function?

#### Tip:
When differentiating or integrating a power series, always carefully adjust the index of summation and confirm the interval of convergence remains unchanged.

Find the power series representation for g centered at 0 by differentiating or integrating the power series for f. Give the interval of convergence for the resulting series. g(x) = 21 / (1 - 21x)^2, f(x) = 1 / (1 - 21x).

Explore how to find the power series representation for g(x) = 21 / (1 - 21x)^2 by differentiating the power series of f(x) = 1 / (1 - 21x). Includes a detailed solution with the interval of convergence for the resulting series. Suitable for college-level calculus students.

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