`0.9 + 0.99 + 0.999 + ……. ` up to `51` terms `= 51 - (1)/(p)(1-(1))/(10^(q))` where `p, q in N` then find the value `(p + q)/(15)`.

Explanation

To solve the problem \(0.9 + 0.99 + 0.999 + \ldots\) up to 51 terms, we can follow these steps:

### Step 1: Rewrite the Series
The series can be expressed as:
\[ S = 0.9 + 0.99 + 0.999 + \ldots \]
This can be rewritten by removing the decimal points:
\[ S = \frac{9}{10} + \frac{99}{100} + \frac{999}{1000} + \ldots \]

### Step 2: Express Each Term
Each term can be expressed in a general form. The \(n\)-th term can be represented as:
\[ \frac{10^n - 1}{10^n} \]
Thus, the series becomes:
\[ S = \sum_{n=1}^{51} \frac{10^n - 1}{10^n} = \sum_{n=1}^{51} \left(1 - \frac{1}{10^n}\right) \]

### Step 3: Separate the Series
We can separate the series into two parts:
\[ S = \sum_{n=1}^{51} 1 - \sum_{n=1}^{51} \frac{1}{10^n} \]
The first part sums to 51:
\[ \sum_{n=1}^{51} 1 = 51 \]

### Step 4: Calculate the Second Series
The second series is a geometric series:
\[ \sum_{n=1}^{51} \frac{1}{10^n} = \frac{\frac{1}{10}(1 - \left(\frac{1}{10}\right)^{51})}{1 - \frac{1}{10}} = \frac{\frac{1}{10}(1 - \frac{1}{10^{51}})}{\frac{9}{10}} = \frac{1 - \frac{1}{10^{51}}}{9} \]

### Step 5: Combine the Results
Now, substituting back into our expression for \(S\):
\[ S = 51 - \frac{1 - \frac{1}{10^{51}}}{9} \]
This simplifies to:
\[ S = 51 - \frac{1}{9} + \frac{1}{9 \cdot 10^{51}} \]
Combining the terms gives:
\[ S = 51 - \frac{1}{9} + \frac{1}{9 \cdot 10^{51}} \]

### Step 6: Final Form
We can express \(S\) in the form given in the question:
\[ S = 51 - \frac{1}{p} \left(1 - \frac{1}{10^q}\right) \]
From our expression, we can identify:
- \(p = 9\)
- \(q = 51\)

### Step 7: Calculate \((p + q)/15\)
Now we find:
\[ p + q = 9 + 51 = 60 \]
Thus:
\[ \frac{p + q}{15} = \frac{60}{15} = 4 \]

### Final Answer
The final answer is:
\[ \boxed{4} \]