If `sum _( k =1) ^(oo) (k^(2))/(3 ^(k))=p/q,` where p and q are relatively prime positive integers. Find the value of `(p-q),`

Answer

To solve the problem, we need to find the sum of the series given by:

\[ S = \sum_{k=1}^{\infty} \frac{k^2}{3^k} \]

We will derive the value of \( S \) step by step.

### Step 1: Identify the series
The series we are dealing with is:

\[ S = \frac{1^2}{3^1} + \frac{2^2}{3^2} + \frac{3^2}{3^3} + \frac{4^2}{3^4} + \ldots \]

### Step 2: Use a known formula for the sum of series
We can use the formula for the sum of the series \( \sum_{k=1}^{\infty} k^2 x^k \), which is given by:

\[ \sum_{k=1}^{\infty} k^2 x^k = \frac{x(1+x)}{(1-x)^3} \]

for \( |x| < 1 \).

### Step 3: Substitute \( x = \frac{1}{3} \)
In our case, we substitute \( x = \frac{1}{3} \):

\[ S = \sum_{k=1}^{\infty} k^2 \left(\frac{1}{3}\right)^k = \frac{\frac{1}{3}(1 + \frac{1}{3})}{\left(1 - \frac{1}{3}\right)^3} \]

### Step 4: Simplify the expression
Now we simplify the expression:

1. Calculate \( 1 + \frac{1}{3} = \frac{4}{3} \).
2. Calculate \( 1 - \frac{1}{3} = \frac{2}{3} \).
3. Therefore, \( \left(1 - \frac{1}{3}\right)^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \).

Now substituting these values back into the formula:

\[ S = \frac{\frac{1}{3} \cdot \frac{4}{3}}{\frac{8}{27}} = \frac{\frac{4}{9}}{\frac{8}{27}} = \frac{4}{9} \cdot \frac{27}{8} = \frac{4 \cdot 27}{9 \cdot 8} \]

### Step 5: Final simplification
Now we simplify \( \frac{4 \cdot 27}{9 \cdot 8} \):

\[ = \frac{108}{72} = \frac{3}{2} \]

### Step 6: Express \( S \) as \( \frac{p}{q} \)
We have found that:

\[ S = \frac{3}{2} \]

Here, \( p = 3 \) and \( q = 2 \).

### Step 7: Calculate \( p - q \)
Now, we need to find \( p - q \):

\[ p - q = 3 - 2 = 1 \]

Thus, the final answer is:

\[ \boxed{1} \]