Key Concepts and Formulas

• Geometric Progression (GP): A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The $k$-th term is given by $a \cdot r^{k-1}$, and the sum of the first $N$ terms is $S_N = a \frac{r^N - 1}{r - 1}$ (for $r \neq 1$).
• Exponent Rules: Key rules include $(a \cdot b)^n = a^n \cdot b^n$ and $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$.
• Prime Factorization: Expressing numbers as a product of their prime factors (e.g., $6 = 2 \cdot 3$) is essential for simplifying expressions involving powers.

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Step-by-Step Solution

Step 1: Analyze the Given Series and Identify the Pattern

The given series is: $S = {6 \over {{3^{12}}}} + {{10} \over {{3^{11}}}} + {{20} \over {{3^{10}}}} + {{40} \over {{3^9}}} + \,\,...\,\, + \,\,{{10240} \over 3}$ Let's denote the terms as $T_1, T_2, T_3, T_4, \dots$. $T_1 = {6 \over {{3^{12}}}}$ $T_2 = {{10} \over {{3^{11}}}}$ $T_3 = {{20} \over {{3^{10}}}}$ $T_4 = {{40} \over {{3^9}}}$

We check the ratio of consecutive terms: $\frac{T_2}{T_1} = \frac{10/3^{11}}{6/3^{12}} = \frac{10}{3^{11}} \times \frac{3^{12}}{6} = \frac{10 \times 3}{6} = 5$. $\frac{T_3}{T_2} = \frac{20/3^{10}}{10/3^{11}} = \frac{20}{3^{10}} \times \frac{3^{11}}{10} = \frac{20 \times 3}{10} = 6$. $\frac{T_4}{T_3} = \frac{40/3^9}{20/3^{10}} = \frac{40}{3^9} \times \frac{3^{10}}{20} = \frac{40 \times 3}{20} = 6$.

Since the ratio is not constant for the first two terms, the entire series is not a single GP. However, from the second term onwards, the common ratio is $r=6$.

Step 2: Manipulate the Series to Form a Standard GP

The terms from $T_2$ onwards form a GP. To make the entire series amenable to GP summation, we can adjust the first term. Let's consider what the first term would be if the GP (with $r=6$) extended backward. This hypothetical first term would be $T_2/6 = (10/3^{11})/6 = 10/(6 \cdot 3^{11}) = 5/(3 \cdot 3^{11}) = 5/3^{12}$.

The actual first term is $6/3^{12}$. We can split it as $6/3^{12} = 1/3^{12} + 5/3^{12}$. Now, we rewrite the series $S$ as: $S = {1 \over {{3^{12}}}} + \left( {{5 \over {{3^{12}}}} + {{10} \over {{3^{11}}}} + {{20} \over {{3^{10}}}} + {{40} \over {{3^9}}} + \,\,...\,\, + \,\,{{10240} \over 3}} \right)$ The terms inside the parenthesis now form a GP, let's call this $S_{GP}$. The first term of this GP is $a = 5/3^{12}$. The common ratio is $r = 6$. The last term of this GP is $L = 10240/3$.

Step 3: Determine the Number of Terms ($N$) in the GP

We use the formula for the $N$-th term of a GP: $L = a \cdot r^{N-1}$. Substituting the values: ${{10240} \over 3} = {{5} \over {{3^{12}}}} \cdot 6^{N-1}$ To solve for $N$, we isolate $6^{N-1}$: $6^{N-1} = {{10240} \over 3} \times {{3^{12}} \over 5}$ $6^{N-1} = {{10240} \over 5} \times {{3^{12}} \over 3}$ $6^{N-1} = 2048 \times 3^{11}$ We know that $2048 = 2^{11}$. So, $6^{N-1} = 2^{11} \times 3^{11}$ Using the exponent rule $(a \cdot b)^n = a^n \cdot b^n$: $6^{N-1} = (2 \cdot 3)^{11}$ $6^{N-1} = 6^{11}$ Equating the exponents, we get $N-1 = 11$, which means $N = 12$. There are 12 terms in the GP $S_{GP}$.

Step 4: Calculate the Sum of the GP ($S_{GP}$)

Using the sum formula $S_N = a \frac{r^N - 1}{r - 1}$: $S_{GP} = {{5} \over {{3^{12}}}} \times \frac{6^{12} - 1}{6 - 1}$ $S_{GP} = {{5} \over {{3^{12}}}} \times \frac{6^{12} - 1}{5}$ The factor of 5 cancels out: $S_{GP} = {{6^{12} - 1} \over {{3^{12}}}}$

Step 5: Calculate the Total Sum ($S$)

The total sum $S$ is the sum of the isolated term and $S_{GP}$: $S = {1 \over {{3^{12}}}} + S_{GP}$ $S = {1 \over {{3^{12}}}} + {{6^{12} - 1} \over {{3^{12}}}}$ Since the denominators are the same, we can combine the numerators: $S = {{1 + (6^{12} - 1)} \over {{3^{12}}}}$ $S = {{6^{12}} \over {{3^{12}}}}$ Using the exponent rule $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$: $S = \left( {6 \over 3} \right)^{12}$ $S = 2^{12}$

Step 6: Express the Sum in the Required Form and Find $m \cdot n$

We are given that $S = 2^n \cdot m$, where $m$ is an odd integer. We found $S = 2^{12}$. Comparing $2^{12}$ with $2^n \cdot m$: We can set $n = 12$ and $m = 1$. Since $m=1$ is indeed an odd integer, this form is correct.

Finally, we need to calculate $m \cdot n$: $m \cdot n = 1 \cdot 12 = 12$

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Common Mistakes & Tips

• Incorrectly Identifying the GP: Always verify the common ratio for at least three consecutive terms to confirm if it's a GP.
• Algebraic Errors with Exponents: Be meticulous when manipulating terms with exponents, especially when combining or dividing terms. Prime factorization of bases (like $6=2 \times 3$) is a powerful tool.
• Forgetting the "m is odd" Condition: Ensure that the factor $m$ in the final form $2^n \cdot m$ is an odd integer. If it's not, you might need to extract all factors of 2 from it.

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Summary

The given series was analyzed to identify that terms from the second one onwards form a Geometric Progression with a common ratio of 6. By strategically splitting the first term, the entire series was expressed as a sum involving a single term and a GP. The number of terms in the GP was determined by solving an exponential equation. The sum of the GP was calculated, and then added to the isolated term to find the total sum of the series. The total sum was simplified to $2^{12}$, which was then expressed in the form $2^n \cdot m$ with $m$ being odd. Finally, the product $m \cdot n$ was computed.

The final answer is \boxed{12}.