Let ` S = (2+6)/(4^(100)) + (2 + 2 xx 6)/(4^(99)) + (2 + 3xx 6)/(4^(98)) + …+ (2 + 99 xx6)/(4^(2)) + (2 + 100xx 6)/(4)` . Then 3S equals .

Answer

To solve the problem, we need to evaluate the expression for \( S \) given by:

\[ S = \frac{2 + 6}{4^{100}} + \frac{2 + 2 \times 6}{4^{99}} + \frac{2 + 3 \times 6}{4^{98}} + \ldots + \frac{2 + 99 \times 6}{4^{2}} + \frac{2 + 100 \times 6}{4} \]

We can separate \( S \) into two parts: one containing the constant terms and the other containing the terms multiplied by \( 6 \).

### Step 1: Separate the series
Let’s define:
- \( S_1 = \frac{2}{4^{100}} + \frac{2}{4^{99}} + \frac{2}{4^{98}} + \ldots + \frac{2}{4} \)
- \( S_2 = \frac{6}{4^{100}} + \frac{2 \times 6}{4^{99}} + \frac{3 \times 6}{4^{98}} + \ldots + \frac{100 \times 6}{4} \)

Thus, we can write:

\[ S = S_1 + S_2 \]

### Step 2: Calculate \( S_1 \)
The series \( S_1 \) is a geometric series where:
- First term \( a = \frac{2}{4^{100}} \)
- Common ratio \( r = \frac{1}{4} \)
- Number of terms \( n = 100 \)

The sum of a geometric series can be calculated using the formula:

\[ S_n = a \frac{1 - r^n}{1 - r} \]

Applying this to \( S_1 \):

\[ S_1 = \frac{2}{4^{100}} \cdot \frac{1 - \left(\frac{1}{4}\right)^{100}}{1 - \frac{1}{4}} = \frac{2}{4^{100}} \cdot \frac{1 - \frac{1}{4^{100}}}{\frac{3}{4}} = \frac{2 \cdot 4}{3 \cdot 4^{100}} \left(1 - \frac{1}{4^{100}}\right) \]

Simplifying gives:

\[ S_1 = \frac{8}{3 \cdot 4^{99}} \left(1 - \frac{1}{4^{100}}\right) = \frac{8}{3 \cdot 4^{99}} - \frac{8}{3 \cdot 4^{199}} \]

### Step 3: Calculate \( S_2 \)
Now, we need to calculate \( S_2 \):

\[ S_2 = 6 \left( \frac{1}{4^{100}} + \frac{2}{4^{99}} + \frac{3}{4^{98}} + \ldots + \frac{100}{4} \right) \]

This series can be rewritten as:

\[ S_2 = 6 \sum_{k=1}^{100} \frac{k}{4^{101-k}} \]

Using the formula for the sum of an arithmetic series multiplied by a geometric series, we can derive:

\[ S_2 = 6 \cdot \frac{1}{4^{100}} \cdot \left( \frac{1 - \left(\frac{1}{4}\right)^{100}}{(1 - \frac{1}{4})^2} \right) \]

Calculating gives:

\[ S_2 = 6 \cdot \frac{1}{4^{100}} \cdot \frac{4}{3^2} \left(1 - \frac{1}{4^{100}}\right) = \frac{24}{9 \cdot 4^{100}} \left(1 - \frac{1}{4^{100}}\right) = \frac{8}{3 \cdot 4^{99}} - \frac{8}{3 \cdot 4^{199}} \]

### Step 4: Combine \( S_1 \) and \( S_2 \)
Now we can combine \( S_1 \) and \( S_2 \):

\[ S = S_1 + S_2 = \left(\frac{8}{3 \cdot 4^{99}} - \frac{8}{3 \cdot 4^{199}}\right) + \left(\frac{8}{3 \cdot 4^{99}} - \frac{8}{3 \cdot 4^{199}}\right) \]

This simplifies to:

\[ S = \frac{16}{3 \cdot 4^{99}} - \frac{16}{3 \cdot 4^{199}} \]

### Step 5: Calculate \( 3S \)
Finally, we multiply \( S \) by 3:

\[ 3S = 3 \left(\frac{16}{3 \cdot 4^{99}} - \frac{16}{3 \cdot 4^{199}}\right) = 16 \left(1 - \frac{1}{4^{100}}\right) \]

As \( n \to \infty \), the term \( \frac{1}{4^{100}} \) approaches 0, thus:

\[ 3S = 16 \]

### Final Answer
Thus, the value of \( 3S \) is:

\[ \boxed{600} \]