If the sum to infinity of the series: `3+(3+d).1/4+(3+2d).1/(4^(2))+`………. Is `4 8/9` find d. Also name the series.

Answer

To solve the problem, we need to find the value of \( d \) in the given series and identify the type of series. The series is:

\[ 3 + (3 + d) \cdot \frac{1}{4} + (3 + 2d) \cdot \frac{1}{4^2} + \ldots \]

We are given that the sum to infinity of this series is \( 4 \frac{8}{9} \) or \( \frac{44}{9} \).

### Step 1: Define the sum of the series

Let \( S \) be the sum of the series:

\[ S = 3 + (3 + d) \cdot \frac{1}{4} + (3 + 2d) \cdot \frac{1}{4^2} + \ldots \]

### Step 2: Write the series in a more manageable form

We can express the series in a more structured way. The series can be rewritten as:

\[ S = 3 + \sum_{n=0}^{\infty} (3 + nd) \cdot \left(\frac{1}{4}\right)^{n+1} \]

### Step 3: Factor out the common terms

We can separate the constant and variable parts:

\[ S = 3 + \sum_{n=0}^{\infty} 3 \cdot \left(\frac{1}{4}\right)^{n+1} + \sum_{n=0}^{\infty} nd \cdot \left(\frac{1}{4}\right)^{n+1} \]

### Step 4: Calculate the first sum

The first sum is a geometric series:

\[ \sum_{n=0}^{\infty} 3 \cdot \left(\frac{1}{4}\right)^{n+1} = 3 \cdot \frac{\frac{1}{4}}{1 - \frac{1}{4}} = 3 \cdot \frac{\frac{1}{4}}{\frac{3}{4}} = 3 \cdot \frac{1}{3} = 1 \]

### Step 5: Calculate the second sum

The second sum can be calculated using the formula for the sum of an infinite series involving \( n \):

\[ \sum_{n=0}^{\infty} n x^n = \frac{x}{(1-x)^2} \]

For our case, \( x = \frac{1}{4} \):

\[ \sum_{n=0}^{\infty} n \cdot \left(\frac{1}{4}\right)^{n+1} = \frac{\frac{1}{4}}{\left(1 - \frac{1}{4}\right)^2} = \frac{\frac{1}{4}}{\left(\frac{3}{4}\right)^2} = \frac{\frac{1}{4}}{\frac{9}{16}} = \frac{16}{36} = \frac{4}{9} \]

Thus, the second sum becomes:

\[ d \cdot \frac{4}{9} \]

### Step 6: Combine the results

Now we can combine everything:

\[ S = 3 + 1 + d \cdot \frac{4}{9} = 4 + d \cdot \frac{4}{9} \]

### Step 7: Set the equation equal to the given sum

We know that \( S = \frac{44}{9} \):

\[ 4 + d \cdot \frac{4}{9} = \frac{44}{9} \]

### Step 8: Solve for \( d \)

Convert \( 4 \) to a fraction with a denominator of \( 9 \):

\[ \frac{36}{9} + d \cdot \frac{4}{9} = \frac{44}{9} \]

Subtract \( \frac{36}{9} \) from both sides:

\[ d \cdot \frac{4}{9} = \frac{44}{9} - \frac{36}{9} = \frac{8}{9} \]

Multiply both sides by \( \frac{9}{4} \):

\[ d = \frac{8}{9} \cdot \frac{9}{4} = 2 \]

### Step 9: Conclusion

The value of \( d \) is \( 2 \).

### Step 10: Identify the series

The series can be identified as an arithmetic-geometric series, where the terms are generated by adding an arithmetic sequence to a geometric sequence.

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