`1+ 4x^(2)+7x^(4)+...`

Answer

To find the sum of the series \( S = 1 + 4x^2 + 7x^4 + \ldots \), we can follow these steps:

### Step 1: Define the series
Let \( S = 1 + 4x^2 + 7x^4 + \ldots \)

### Step 2: Multiply the series by \( x^2 \)
Now, multiply the entire series \( S \) by \( x^2 \):
\[ S x^2 = x^2 + 4x^4 + 7x^6 + \ldots \]

### Step 3: Subtract the two equations
Now, subtract the equation obtained in Step 2 from the original series:
\[ S - S x^2 = (1 + 4x^2 + 7x^4 + \ldots) - (x^2 + 4x^4 + 7x^6 + \ldots) \]
This simplifies to:
\[ S(1 - x^2) = 1 + 3x^2 + 3x^4 + 3x^6 + \ldots \]

### Step 4: Identify the right-hand side as a geometric series
The right-hand side \( 1 + 3x^2 + 3x^4 + 3x^6 + \ldots \) can be factored:
\[ 1 + 3x^2(1 + x^2 + x^4 + \ldots) \]
The series \( 1 + x^2 + x^4 + \ldots \) is a geometric series with first term \( 1 \) and common ratio \( x^2 \). The sum of this series is:
\[ \frac{1}{1 - x^2} \]
Thus, we can rewrite the right-hand side:
\[ 1 + 3x^2 \cdot \frac{1}{1 - x^2} = 1 + \frac{3x^2}{1 - x^2} \]

### Step 5: Combine and simplify
Now we have:
\[ S(1 - x^2) = 1 + \frac{3x^2}{1 - x^2} \]
To combine the terms on the right-hand side, we can express \( 1 \) as \( \frac{1 - x^2}{1 - x^2} \):
\[ S(1 - x^2) = \frac{(1 - x^2) + 3x^2}{1 - x^2} = \frac{1 + 2x^2}{1 - x^2} \]

### Step 6: Solve for \( S \)
Now, divide both sides by \( 1 - x^2 \):
\[ S = \frac{1 + 2x^2}{(1 - x^2)^2} \]

### Final Answer
Thus, the sum of the series is:
\[ S = \frac{1 + 2x^2}{(1 - x^2)^2} \]

---