If `3+1/4(3+p)+1/(4^2)(3+2p)+... + oo=8`, then p

Options

A. 1
B. 5
C. 7
D. 9

Answer

To solve the equation \(3 + \frac{1}{4}(3 + p) + \frac{1}{4^2}(3 + 2p) + \ldots = 8\), we will break it down step by step.

### Step 1: Rewrite the Series
We can express the series as follows:
\[ S = 3 + \frac{1}{4}(3 + p) + \frac{1}{4^2}(3 + 2p) + \ldots \]
This can be split into two separate series:
\[ S = 3 + \left(\frac{3}{4} + \frac{3}{4^2} + \ldots\right) + \left(\frac{p}{4} + \frac{2p}{4^2} + \ldots\right) \]

### Step 2: Calculate the First Series
The first series is a geometric series:
\[ \frac{3}{4} + \frac{3}{4^2} + \ldots \]
The first term \(a = \frac{3}{4}\) and the common ratio \(r = \frac{1}{4}\). The sum of an infinite geometric series is given by:
\[ \text{Sum} = \frac{a}{1 - r} = \frac{\frac{3}{4}}{1 - \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{3}{4}} = 1 \]
Thus, the sum of the first series is 1.

### Step 3: Calculate the Second Series
The second series is:
\[ \frac{p}{4} + \frac{2p}{4^2} + \ldots \]
We can factor out \(p\):
\[ p\left(\frac{1}{4} + \frac{2}{4^2} + \ldots\right) \]
Let \(k = \frac{1}{4} + \frac{2}{4^2} + \frac{3}{4^3} + \ldots\). This can be expressed as:
\[ k = \sum_{n=1}^{\infty} \frac{n}{4^n} \]
Using the formula for the sum of \(n x^n\):
\[ \sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2} \quad \text{for } |x| < 1 \]
Here, \(x = \frac{1}{4}\):
\[ k = \frac{\frac{1}{4}}{\left(1 - \frac{1}{4}\right)^2} = \frac{\frac{1}{4}}{\left(\frac{3}{4}\right)^2} = \frac{\frac{1}{4}}{\frac{9}{16}} = \frac{4}{9} \]
Thus, the second series becomes:
\[ p \cdot \frac{4}{9} \]

### Step 4: Combine the Results
Now we can combine the results:
\[ S = 3 + 1 + p \cdot \frac{4}{9} = 4 + \frac{4p}{9} \]
We are given that \(S = 8\):
\[ 4 + \frac{4p}{9} = 8 \]

### Step 5: Solve for \(p\)
Subtract 4 from both sides:
\[ \frac{4p}{9} = 4 \]
Multiply both sides by 9:
\[ 4p = 36 \]
Divide by 4:
\[ p = 9 \]

### Final Answer
Thus, the value of \(p\) is:
\[ \boxed{9} \]