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 of finding the sum of the series \( S = \sum_{k=1}^{\infty} \frac{k^2}{3^k} \) and expressing it in the form \( \frac{p}{q} \), where \( p \) and \( q \) are relatively prime positive integers, we can follow these steps:

### Step 1: Define the Sum
Let:
\[ S = \sum_{k=1}^{\infty} \frac{k^2}{3^k} \]

### Step 2: Create a Related Series
To manipulate this series, we can consider the series:
\[ \frac{S}{3} = \sum_{k=1}^{\infty} \frac{k^2}{3^{k+1}} = \sum_{k=2}^{\infty} \frac{(k-1)^2}{3^k} \]
This shifts the index of summation.

### Step 3: Subtract the Two Series
Now, we can subtract \( \frac{S}{3} \) from \( S \):
\[ S - \frac{S}{3} = \sum_{k=1}^{\infty} \frac{k^2}{3^k} - \sum_{k=2}^{\infty} \frac{(k-1)^2}{3^k} \]
The left-hand side simplifies to:
\[ \frac{2S}{3} \]
The right-hand side becomes:
\[ \frac{1^2}{3^1} + \sum_{k=2}^{\infty} \left( \frac{k^2}{3^k} - \frac{(k-1)^2}{3^k} \right) \]
The term \( \frac{1^2}{3^1} = \frac{1}{3} \).

### Step 4: Simplify the Right-Hand Side
For \( k \geq 2 \):
\[ k^2 - (k-1)^2 = k^2 - (k^2 - 2k + 1) = 2k - 1 \]
Thus, we have:
\[ \sum_{k=2}^{\infty} \frac{2k - 1}{3^k} \]
This can be split into two separate sums:
\[ 2\sum_{k=2}^{\infty} \frac{k}{3^k} - \sum_{k=2}^{\infty} \frac{1}{3^k} \]

### Step 5: Evaluate the Sums
The sum \( \sum_{k=2}^{\infty} \frac{1}{3^k} \) is a geometric series:
\[ \sum_{k=2}^{\infty} \frac{1}{3^k} = \frac{\frac{1}{3^2}}{1 - \frac{1}{3}} = \frac{\frac{1}{9}}{\frac{2}{3}} = \frac{1}{6} \]

Next, we need to find \( \sum_{k=1}^{\infty} \frac{k}{3^k} \):
\[ \sum_{k=1}^{\infty} \frac{k}{3^k} = \frac{3}{(3-1)^2} = \frac{3}{4} \]
Thus,
\[ \sum_{k=2}^{\infty} \frac{k}{3^k} = \sum_{k=1}^{\infty} \frac{k}{3^k} - \frac{1}{3} = \frac{3}{4} - \frac{1}{3} = \frac{9 - 4}{12} = \frac{5}{12} \]

### Step 6: Substitute Back
Now substituting back into our equation:
\[ \frac{2S}{3} = \frac{1}{3} + 2 \cdot \frac{5}{12} - \frac{1}{6} \]
Calculating the right-hand side:
\[ \frac{1}{3} + \frac{10}{12} - \frac{2}{12} = \frac{1}{3} + \frac{8}{12} = \frac{1}{3} + \frac{2}{3} = 1 \]

### Step 7: Solve for S
Now we have:
\[ \frac{2S}{3} = 1 \implies 2S = 3 \implies S = \frac{3}{2} \]

### Step 8: Identify p and q
In the form \( \frac{p}{q} \), we have \( p = 3 \) and \( q = 2 \). Since 3 and 2 are relatively prime, we can find:
\[ p + q = 3 + 2 = 5 \]

### Final Answer
The final answer is:
\[ \boxed{5} \]