If 6/3¹² + 10/3¹¹ + 20/3¹⁰ + 40/3⁹ + ... + 10240/3 = 2ⁿ · m, where m is odd, then m · n is equal to

Q: Wait, why isn't the whole series a G.P.? The terms look like they are growing.

That is a brilliant observation! While the terms are indeed growing, a geometric progression requires a *constant* ratio between *every* consecutive pair. If you calculate the ratio of the second term to the first, you get $5$. However, if you calculate the ratio of the third term to the second, you get $6$. Because the ratio changes, the first term is an outlier. Always check the ratio for at least two pairs before assuming a G.P.!

Q: How do I know how many terms are in the G.P. part?

Look at the powers of $3$ in the denominator. They range from $11$ down to $1$. If you list these exponents: $11, 10, 9, \dots, 1$, you are counting integers from $1$ to $11$. Therefore, there are exactly $11$ terms in the geometric progression. Always map the indices to the exponents to avoid off-by-one errors.

Q: I got a messy fraction when I added the outlier. Did I do something wrong?

Don't panic! When you combine the outlier $\frac{6}{3^{12}}$ with the G.P. sum, you will need a common denominator of $3^{12}$. When you multiply the G.P. sum by $\frac{3}{3}$, the numerator will become $6 + 6(6^{11} - 1)$. Notice that the $6$ and the $-6$ inside the expansion will cancel out perfectly. It is designed to be elegant, so trust the algebra!

Question

If $\frac{6}{3^{12}} + \frac{10}{3^{11}} + \frac{20}{3^{10}} + \frac{40}{3^9} + \dots + \frac{10240}{3} = 2^n \cdot m$, where $m$ is odd, then $m \cdot n$ is equal to ________.

Sigma Percentile · Accepted Answer

### The Art of Skepticism: Unmasking the Imposter

Welcome, future engineer. Today, we are not just solving a series; we are unmasking a mathematical imposter.

When you look at the series

$$S = \frac{6}{3^{12}} + \frac{10}{3^{11}} + \frac{20}{3^{10}} + \dots + \frac{10240}{3}$$

your brain naturally wants to categorize it. It looks like a geometric progression, but in the high-stakes arena of JEE Advanced, the first step is always skepticism. We never assume; we verify.

### Phase 1: The Detective Work

Let us test the ratio. If this were a perfect G.P., the ratio between any two consecutive terms would be constant.

Let's check the ratio of the second term to the first:

$$\frac{10/3^{11}}{6/3^{12}} = \frac{10}{6} \cdot 3 = 5$$

Now, let's check the ratio of the third term to the second:

$$\frac{20/3^{10}}{10/3^{11}} = \frac{20}{10} \cdot 3 = 6$$

Do you see it? The ratio shifts from $5$ to $6$. This is the 'Aha!' moment.

The first term, $\frac{6}{3^{12}}$, is an outlier. It does not belong to the geometric progression that follows. By isolating it, we clear the path to use our standard tools.

### Phase 2: The Calculation

Now that we have identified the G.P. starting from the second term, let's define our parameters. The first term of our G.P. is $a = \frac{10}{3^{11}}$ and the common ratio is $r = 6$.

To find the number of terms $n$, we look at the powers of $3$ in the denominator, which run from $11$ down to $1$. This gives us $n = 11$ terms.

We invoke the sum formula:

$$S_{GP} = a \frac{r^n - 1}{r - 1}$$

Substituting our values, we get:

$$S_{GP} = \frac{10}{3^{11}} \cdot \frac{6^{11} - 1}{6 - 1}$$

The denominator $6 - 1$ is $5$, and $\frac{10}{5}$ simplifies beautifully to $2$. Thus, our G.P. sum is:

$$S_{GP} = \frac{2(6^{11} - 1)}{3^{11}}$$

### Phase 3: The Grand Finale

We are almost there. We must add our outlier back to the G.P. sum:

$$S = \frac{6}{3^{12}} + \frac{2(6^{11} - 1)}{3^{11}}$$

To add these, we need a common denominator of $3^{12}$. We multiply the second term by $\frac{3}{3}$, which gives us:

$$\frac{6 + 6(6^{11} - 1)}{3^{12}}$$

Now, watch the magic of algebra. Expanding the numerator, we get $6 + 6^{12} - 6$. The $6$ and $-6$ cancel out, leaving us with just $6^{12}$ in the numerator.

We are left with:

$$S = \frac{6^{12}}{3^{12}} = \left(\frac{6}{3}\right)^{12} = 2^{12}$$

### Conclusion

We are given $S = 2^n \cdot m$, where $m$ is odd. We found $S = 2^{12} \cdot 1$.

By comparison, $n = 12$ and $m = 1$. The product **$m \cdot n = 12$**.

This problem teaches us that even when a series looks chaotic, there is an underlying order waiting to be revealed. Keep your eyes sharp, trust your algebraic foundations, and always look for the symmetry.

Animated Solution for Mathematics - Sequence and Series: If $\frac{6}{3 ^{12}} + \frac{10}{3 ^{11}} + \frac{20}{3 ^{10}} + \frac{40}{3 ^{9}} + \dots + \frac{10240}{3} = 2^{n} \cdot m$ , where $m$ is odd, then $m \cdot n$ is equal to ________.

Enter Numerical Value:

Visualized Solution

Analyze the Series Structure

Identify the Outlier Term

Define the G.P. Parameters

Count the Number of Terms

Apply the G.P. Sum Formula

Substitute and Simplify Denominator

Finalize G.P. Sum Expression

Combine with the First Term

Create a Common Denominator

Expand and Cancel Terms

Simplify Using Exponent Laws

Compare and Find $m$ and $n$

Final Calculation

The Sigma Insight: Geometric Progression (G.P.)

The Art of Skepticism

Phase 1

Phase 2

Phase 3

Conclusion

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