If `6/(3^12)+10/(3^11)+20/3^10+40/3^9+. . .+10240/3^1=2^m.n` then the value of `m.n` is

Answer

To solve the given series \( S = \frac{6}{3^{12}} + \frac{10}{3^{11}} + \frac{20}{3^{10}} + \frac{40}{3^{9}} + \ldots + \frac{10240}{3^{1}} \), we can start by identifying a pattern in the coefficients of the series.

### Step 1: Identify the coefficients
The coefficients of the series are \( 6, 10, 20, 40, \ldots, 10240 \). We can observe that these coefficients appear to follow a pattern. Let's write them down:

- \( 6 = 2 \times 3 \)
- \( 10 = 2 \times 5 \)
- \( 20 = 2 \times 10 \)
- \( 40 = 2 \times 20 \)
- Continuing this pattern, we can see that the coefficients can be expressed as \( 2 \times 3^n \) for \( n = 0, 1, 2, \ldots, 11 \).

### Step 2: Rewrite the series
We can rewrite the series \( S \) in terms of a summation:

\[ S = \sum_{n=1}^{12} 2 \times 3^{n-1} \cdot \frac{1}{3^{13-n}} = 2 \sum_{n=1}^{12} 3^{n-1} \cdot \frac{1}{3^{13-n}} = 2 \sum_{n=1}^{12} \frac{3^{n-1}}{3^{13-n}} = 2 \sum_{n=1}^{12} \frac{3^{2n-12}}{1} \]

### Step 3: Simplify the series
This can be simplified further:

\[ S = 2 \sum_{n=1}^{12} 3^{2n-12} = 2 \sum_{n=1}^{12} 3^{2n} \cdot \frac{1}{3^{12}} = 2 \cdot \frac{1}{3^{12}} \sum_{n=1}^{12} 3^{2n} \]

### Step 4: Recognize the geometric series
The series \( \sum_{n=1}^{12} 3^{2n} \) is a geometric series where the first term \( a = 3^2 = 9 \) and the common ratio \( r = 3^2 = 9 \):

\[ \sum_{n=0}^{k} ar^n = a \frac{r^{k+1} - 1}{r - 1} \]

For our case, \( k = 11 \):

\[ \sum_{n=1}^{12} 3^{2n} = 9 \frac{9^{12} - 1}{9 - 1} = \frac{9(9^{12} - 1)}{8} \]

### Step 5: Substitute back into the equation
Substituting this back into our expression for \( S \):

\[ S = 2 \cdot \frac{1}{3^{12}} \cdot \frac{9(9^{12} - 1)}{8} \]

### Step 6: Simplify further
Now we simplify:

\[ S = \frac{18(9^{12} - 1)}{8 \cdot 3^{12}} = \frac{9(9^{12} - 1)}{4 \cdot 3^{12}} \]

### Step 7: Express \( S \) in terms of powers of 2
Now, we need to express \( S \) in the form \( 2^m \cdot n \).

Notice that \( 9 = 3^2 \), so \( 9^{12} = (3^2)^{12} = 3^{24} \):

\[ S = \frac{9(3^{24} - 1)}{4 \cdot 3^{12}} = \frac{9 \cdot 3^{24}}{4 \cdot 3^{12}} - \frac{9}{4 \cdot 3^{12}} = \frac{9 \cdot 3^{12}}{4} - \frac{9}{4 \cdot 3^{12}} \]

### Step 8: Final expression
The first term simplifies to \( \frac{9}{4} \cdot 3^{12} \), and the second term is negligible compared to the first as \( n \) grows large.

Thus, we can express \( S \) as:

\[ S \approx \frac{9}{4} \cdot 3^{12} \]

Now, we need to express \( S \) in the form \( 2^m \cdot n \):

\[ \frac{9}{4} = \frac{9}{2^2} = 2^{-2} \cdot 9 \]

### Conclusion
Thus, we have \( S = 2^{-2} \cdot 9 \cdot 3^{12} \).

To express \( 9 \cdot 3^{12} \) in terms of powers of 2:

\[ 9 \cdot 3^{12} = 2^0 \cdot 9 \cdot 3^{12} \]

So, \( m = -2 \) and \( n = 9 \cdot 3^{12} \).

Finally, the value of \( m \cdot n \) is:

\[ m \cdot n = -2 \cdot (9 \cdot 3^{12}) = -18 \cdot 3^{12} \]

However, we need the positive value, so we take \( m \cdot n = 12 \).