Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

This problem involves an infinite series where an arithmetic progression is combined with powers of two, a structure that simplifies efficiently using the method of shifting and subtraction. If we define the given sum as $S = \frac{a_1}{2} + \frac{a_2}{2^2} + \frac{a_3}{2^3} + \dots = 4$, we can isolate the common difference by evaluating $\frac{S}{2}$ and subtracting it from the original series. This operation yields $\frac{S}{2} = \frac{a_1}{2} + \frac{a_2 - a_1}{2^2} + \frac{a_3 - a_2}{2^3} + \dots$, where the numerator of each term after the first represents the constant common difference $d$ of the arithmetic progression.

With the common difference $d$ substituted into the series, the expression becomes $\frac{S}{2} = \frac{a_1}{2} + \frac{d}{2^2} + \frac{d}{2^3} + \dots$. The infinite portion of this sum is a geometric series with a first term of $\frac{d}{4}$ and a common ratio of $\frac{1}{2}$, which sums to $\frac{d/4}{1 - 1/2}$, simplifying perfectly to $\frac{d}{2}$. Consequently, the equation simplifies to $\frac{S}{2} = \frac{a_1}{2} + \frac{d}{2}$, which implies $S = a_1 + d$.

Since the second term of an arithmetic progression is defined as $a_2 = a_1 + d$, we can conclude that the sum $S$ is equivalent to $a_2$. Given that the problem states $S = 4$, it follows that $a_2 = 4$. Therefore, calculating the value requested in the problem, $4a_2$, results in $4 \times 4 = 16$.

The relationship between the numbers 12 and 19 is best understood through the quadratic equation $x^2 - 31x + 228 = 0$, which implies that $31 - x = \frac{228}{x}$ for either root. Expanding the numerator $31 \alpha_9 - \alpha_{10}$ using the definition $\alpha_n = 19^n - 12^n$ results in $31(19^9 - 12^9) - (19^{10} - 12^{10})$. Rearranging these terms by grouping the powers of 19 and 12 yields $19^9(31 - 19) - 12^9(31 - 12)$, which simplifies to $19^9(12) - 12^9(19)$. Factoring out the product $19 \cdot 12$, which is 228, results in $228(19^8 - 12^8)$ or $228 \alpha_8$, and dividing this by the denominator $57 \alpha_8$ gives 4.

The Arithmetic Mean-Geometric Mean (AM-GM) inequality provides the most direct method to find the minimum value of a sum when given a product constraint. To apply this principle, express the target expression $3x + 2y$ as a sum of five individual variables by grouping three instances of $x$ and two instances of $y$, such that $\frac{x+x+x+y+y}{5} \ge \sqrt[5]{x^3y^2}$. Substituting the known value $x^3y^2 = 2^{15}$ into this inequality yields $\frac{3x+2y}{5} \ge \sqrt[5]{2^{15}}$, which simplifies to $\frac{3x+2y}{5} \ge 2^3$. Multiplying both sides by 5 confirms that the least possible value for $3x + 2y$ is 40.

The sum of an infinite geometric progression with first term $a$ and common ratio $r$ is defined as $\frac{a}{1-r}=5$. The sum of the first five terms of this progression is expressed as $\frac{a(1-r^5)}{1-r} = \frac{98}{25}$. By substituting the known value of the infinite sum $\frac{a}{1-r}=5$ into this equation, we obtain $5(1-r^5) = \frac{98}{25}$, which simplifies to $1-r^5 = \frac{98}{125}$, and further confirms that $r^5 = \frac{27}{125}$.

For the arithmetic progression with first term $A = 10ar$ and common difference $D = 10(ar)^2$, the sum of the first 21 terms is calculated using the formula $S_n = \frac{n}{2}[2A + (n-1)D]$. Substituting the given values for this specific progression yields $S_{21} = \frac{21}{2}[2(10ar) + (21-1)(10(ar)^2)]$. This expression simplifies to $\frac{21}{2}[20ar + 200(ar)^2]$, which reduces to $21[10ar + 100(ar)^2]$.

The eleventh term of an arithmetic progression is given by $a_n = A + (n-1)D$. For this progression, $a_{11} = 10ar + (11-1)(10(ar)^2)$, which is $10ar + 10(10(ar)^2)$, or $10ar + 100(ar)^2$. By comparing the derived sum $S_{21} = 21[10ar + 100(ar)^2]$ with the eleventh term $a_{11}$, it is evident that the sum is equal to $21 a_{11}$.

The sequence is defined by the relation $a_{n+2}=2a_{n+1}-a_n+1$ with initial values $a_0=0$ and $a_1=0$. Calculating the next terms in the sequence, we obtain $a_2=2(0)-0+1=1$, $a_3=2(1)-0+1=3$, $a_4=2(3)-1+1=6$, and $a_5=2(6)-3+1=10$. The sum to be evaluated is $S = \frac{1}{7^2} + \frac{3}{7^3} + \frac{6}{7^4} + \frac{10}{7^5} + \dots$.

The arithmetic-geometric progression subtraction method allows us to find the total value of this series. Multiplying the entire sum $S$ by $\frac{1}{7}$ results in $\frac{1}{7}S = \frac{1}{7^3} + \frac{3}{7^4} + \frac{6}{7^5} + \dots$. Subtracting this shifted series from the original $S$ leaves $\frac{6}{7}S = \frac{1}{7^2} + \frac{2}{7^3} + \frac{3}{7^4} + \frac{4}{7^5} + \dots$. Applying the subtraction process again, we multiply this result by $\frac{1}{7}$ once more, yielding $\frac{6}{49}S = \frac{1}{7^3} + \frac{2}{7^4} + \frac{3}{7^5} + \dots$. Subtracting this second shifted series from the result of the first subtraction gives $(\frac{6}{7} - \frac{6}{49})S = \frac{1}{7^2} + \frac{1}{7^3} + \frac{1}{7^4} + \dots$.

The right side of this equation is a simple infinite geometric series with a starting term of $\frac{1}{49}$ and a common ratio of $\frac{1}{7}$. The sum of this geometric series is $\frac{1/49}{1-1/7}$, which simplifies to $\frac{1/49}{6/7}$ or $\frac{1}{42}$. Simultaneously, the left side simplifies to $\frac{36}{49}S$. Equating these two values, we have $\frac{36}{49}S = \frac{1}{42}$, which means $S = \frac{1}{42} \cdot \frac{49}{36}$. After performing the multiplication and reduction, the final value of the sum is $\frac{7}{216}$.

This type of problem involves an arithmetico-geometric series, a sequence where each term is the product of terms from an arithmetic progression and a geometric progression. To evaluate such a sum, we utilize the method of differences, which involves subtracting shifted versions of the series from itself to reduce the polynomial part of the numerator to a constant, thereby revealing a standard geometric series.

Starting with the given series $S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+\dots$, we multiply both sides by $\frac{1}{7}$ to obtain $\frac{1}{7}S=\frac{2}{7}+\frac{6}{7^2}+\frac{12}{7^3}+\frac{20}{7^4}+\dots$. Subtracting this second equation from the original $S$ effectively reduces the numerators, yielding $\frac{6}{7}S=2+\frac{4}{7}+\frac{6}{7^2}+\frac{8}{7^3}+\frac{10}{7^4}+\dots$. By applying this subtraction process a second time, we multiply the new series by $\frac{1}{7}$, resulting in $\frac{6}{7^2}S=\frac{2}{7}+\frac{4}{7^2}+\frac{6}{7^3}+\frac{8}{7^4}+\dots$.

Subtracting this second result from $\frac{6}{7}S$ leaves us with $\left(\frac{6}{7} - \frac{6}{7^2}\right)S = 2 + \frac{2}{7} + \frac{2}{7^2} + \frac{2}{7^3} + \dots$. Factoring the left side produces $\frac{36}{49}S$, while the right side is a geometric progression with a first term of $2$ and a common ratio of $\frac{1}{7}$, which sums to $2 + \frac{2/7}{1-1/7} = 2 + \frac{1}{3} = \frac{7}{3}$. Solving for $S$ gives $S = \frac{7}{3} \cdot \frac{49}{36} = \frac{343}{108}$, and multiplying by $4$ produces $4S = \frac{343}{27}$, which is equivalent to $\left(\frac{7}{3}\right)^3$.

The behavior of these infinite series depends primarily on the term $(3+(-1)^n)^n$, which changes based on whether the index $n$ is odd or even. When $n$ is odd, the base becomes $(3-1)=2$, making the terms involve powers of $2$, whereas when $n$ is even, the base becomes $(3+1)=4$, making the terms involve powers of $4$. Separating the odd and even indexed terms allows us to express both $A$ and $B$ as the sum of two distinct convergent geometric series.

For the series $A$, the odd-indexed terms follow the sequence $\frac{1}{2} + \frac{1}{8} + \dots$, which sums to $\frac{1/2}{1-1/4} = \frac{2}{3}$, while the even-indexed terms follow $\frac{1}{16} + \frac{1}{16^2} + \dots$, summing to $\frac{1/16}{1-1/16} = \frac{1}{15}$. Combining these gives $A = \frac{10}{15} + \frac{1}{15} = \frac{11}{15}$. For the series $B$, the even-indexed terms remain identical, but the odd-indexed terms are negated because of the $(-1)^n$ factor in the numerator, creating a series $-\frac{1}{2} - \frac{1}{8} - \dots$ that sums to $-\frac{2}{3}$. This results in $B = -\frac{10}{15} + \frac{1}{15} = -\frac{9}{15}$. Dividing these two results, $\frac{A}{B} = \frac{11/15}{-9/15}$, provides the final value of $-\frac{11}{9}$.

An arithmetic progression follows a linear pattern determined by its starting term and common difference, where any term $a_n$ is given by $a_n = a_1 + (n-1)d$. Given the first sequence has $a_1=2$ and $a_{10}=3$, the common difference is $\frac{3-2}{10-1} = \frac{1}{9}$, meaning the fourth term is $a_4 = 2 + 3(\frac{1}{9}) = \frac{7}{3}$.

The second arithmetic progression requires identifying its starting term and common difference from the relationship $a_n b_n = 1$. Since $a_1=2$, we have $b_1 = \frac{1}{2}$; similarly, $a_{10}=3$ implies $b_{10} = \frac{1}{3}$. Calculating the common difference for this second progression gives $\frac{1/3 - 1/2}{10-1} = -\frac{1}{54}$, allowing us to determine $b_4 = \frac{1}{2} + 3(-\frac{1}{54}) = \frac{1}{2} - \frac{1}{18} = \frac{4}{9}$. The final product $a_4 b_4 = \frac{7}{3} \times \frac{4}{9} = \frac{28}{27}$.

When dealing with equations where the reciprocals of the roots form an Arithmetic Progression, one effective technique is to transform the original equations into their reciprocal forms. Given the quadratic equation $x^{2} - 8 a x + 2 a = 0$ with roots $p$ and $r$, substituting $y = \frac{1}{x}$ yields the reciprocal equation $2 a y^{2} - 8 a y + 1 = 0$, whose roots are $\frac{1}{p}$ and $\frac{1}{r}$. Similarly, transforming $x^{2} + 12 b x + 6 b = 0$ with roots $q$ and $s$ produces $6 b y^{2} + 12 b y + 1 = 0$, having roots $\frac{1}{q}$ and $\frac{1}{s}$.

Since $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ are in an Arithmetic Progression, we can represent these four terms as $\alpha - 3\beta, \alpha - \beta, \alpha + \beta, \alpha + 3\beta$. Based on the sum of the roots for the transformed equations, the sum of the first and third terms is $\frac{1}{p} + \frac{1}{r} = (\alpha - 3\beta) + (\alpha + \beta) = 2\alpha - 2\beta = 4$, while the sum of the second and fourth terms is $\frac{1}{q} + \frac{1}{s} = (\alpha - \beta) + (\alpha + 3\beta) = 2\alpha + 2\beta = -2$. Solving this system of equations provides $\alpha = \frac{1}{2}$ and $\beta = -\frac{3}{2}$.

Using the product of the roots for each quadratic equation, we find that the product $\frac{1}{p} \cdot \frac{1}{r} = (\alpha - 3\beta)(\alpha + \beta) = \frac{1}{2a}$. Substituting our values gives $(\frac{1}{2} + \frac{9}{2})(\frac{1}{2} - \frac{3}{2}) = (5)(-1) = -5$, leading to $\frac{1}{a} = -10$. Similarly, the product $\frac{1}{q} \cdot \frac{1}{s} = (\alpha - \beta)(\alpha + 3\beta) = \frac{1}{6b}$ yields $(\frac{1}{2} + \frac{3}{2})(\frac{1}{2} - \frac{9}{2}) = (2)(-4) = -8$, implying $\frac{1}{b} = -48$. Calculating the difference $\frac{1}{a} - \frac{1}{b}$ results in $-10 - (-48) = 38$.

Telescoping series provide an efficient way to evaluate sums by decomposing the general term into simpler fractions that cancel each other out. By applying partial fraction decomposition to the expression $\frac{3}{(4n-1)(4n+3)}$, it transforms into the form $\frac{3}{4} \left( \frac{1}{4n-1} - \frac{1}{4n+3} \right)$, creating the necessary structure for terms to cancel sequentially.

Expanding the summation from $n=1$ up to $n=21$ reveals that every intermediate term vanishes, leaving only the first component of the initial term and the second component of the final term. This simplification reduces the entire expression to $\frac{3}{4} \left( \frac{1}{3} - \frac{1}{87} \right)$. Computing the value within the parentheses yields $\frac{28}{87}$, and multiplying this by $\frac{3}{4}$ results in the final value of $\frac{7}{29}$.

The sum of an infinite geometric series with first term $a = n^2$ and common ratio $r = \frac{1}{(n+1)^2}$ is given by the formula $S_n = \frac{a}{1-r}$. Substituting the values produces $S_n = \frac{n^2}{1 - \frac{1}{(n+1)^2}}$, which simplifies to $\frac{n^2(n+1)^2}{(n+1)^2 - 1} = \frac{n(n+1)^2}{n+2}$. By performing polynomial division, this expression is rewritten as $S_n = n^2 + 1 - \frac{2}{n+2}$.

Substituting this into the required expression yields:
$\frac{1}{26} + \sum_{n=1}^{50} \left(n^2 - n + 2\left(\frac{1}{n+1} - \frac{1}{n+2}\right)\right)$
This sum consists of the sum of the first 50 squares, the negative sum of the first 50 integers, and a telescoping series. Calculating these components results in:
$\frac{50 \times 51 \times 101}{6} - \frac{50 \times 51}{2} + 2\left(\frac{1}{2} - \frac{1}{52}\right)$
Evaluating the arithmetic yields $42925 - 1275 + (1 - \frac{1}{26})$. Including the initial $\frac{1}{26}$ term cancels the fractional part, leaving $42925 - 1275 + 1$, which results in a final value of $41651$.

The core of this problem relies on identifying the algebraic structure of the denominator, which can be factored by recognizing it as a difference of squares. By rewriting $k^4+k^2+1$ as $(k^2+1)^2 - k^2$, we can factor it into $(k^2-k+1)(k^2+k+1)$. This allows us to decompose the general term into partial fractions:

$\frac{k}{k^4+k^2+1} = \frac{1}{2}\left(\frac{1}{k^2-k+1} - \frac{1}{k^2+k+1}\right)$

This expression is particularly useful because it creates a telescoping series. As we sum the terms from $k=1$ to $k=10$, the negative part of each term cancels with the positive part of the subsequent term. For instance, the second part of the $k=1$ term is the negative of the first part of the $k=2$ term. Consequently, all intermediate values cancel out, leaving us with only the first component of the $k=1$ term and the last component of the $k=10$ term.

Calculating these boundaries, the first part at $k=1$ is $\frac{1}{1^2-1+1} = 1$, and the second part at $k=10$ is $\frac{1}{10^2+10+1} = \frac{1}{111}$. The total sum is therefore $\frac{1}{2}\left(1 - \frac{1}{111}\right)$, which simplifies to $\frac{1}{2} \cdot \frac{110}{111} = \frac{55}{111}$. Since $55$ and $111$ are co-prime, we have $m=55$ and $n=111$, giving a final sum of $m+n = 166$.

In a geometric progression where each term follows the rule $A_n = A_1 r^{n-1}$, the product of terms equidistant from the center simplifies based on the constant common ratio. Specifically, the product of the terms $A_1$, $A_3$, $A_5$, and $A_7$ expands to $A_1 \cdot (A_1 r^2) \cdot (A_1 r^4) \cdot (A_1 r^6)$, which simplifies to $(A_1 r^3)^4$. Given that the term $A_1 r^3$ is equivalent to $A_4$, the equation $(A_4)^4 = 1/1296$ allows us to identify the fourth term as $A_4 = 1/6$.

With the fourth term known, the sum $A_2 + A_4 = 7/36$ lets us determine $A_2 = 7/36 - 6/36 = 1/36$. The common ratio squared is found by dividing $A_4$ by $A_2$, resulting in $r^2 = (1/6) / (1/36) = 6$. Applying this factor repeatedly to $A_4$ gives $A_6 = A_4 \cdot r^2 = 1$, $A_8 = A_6 \cdot r^2 = 6$, and $A_{10} = A_8 \cdot r^2 = 36$. The requested sum of these three terms, $1 + 6 + 36$, results in 43.

The series consists of terms where the denominator is a triangular number, defined as the sum of the first $n$ positive integers, $\frac{n(n+1)}{2}$. By expressing the general term $T_n$ as the reciprocal of this sum, we obtain:
$T_n = \frac{2}{n(n+1)}$
Applying partial fraction decomposition allows us to rewrite this general term as:
$T_n = 2\left(\frac{1}{n} - \frac{1}{n+1}\right)$
Summing these terms from $n=1$ to $n=11$ creates a telescoping series where each intermediate term cancels out the preceding one. This leaves only the first and last components, resulting in the calculation:
$S_{11} = 2\left(1 - \frac{1}{12}\right) = 2\left(\frac{11}{12}\right) = \frac{11}{6}$

This series is an Arithmetico-Geometric Progression because each term contains the product of an arithmetic component and a geometric component. Calculating such a sum involves multiplying the expression by the common ratio and subtracting the resulting series from the original to collapse the middle terms. Defining the sum as $S = 1 + 2\cdot 3 + 3\cdot 3^2 + \dots + 10\cdot 3^9$, multiplying by $3$ gives $3S = 1\cdot 3 + 2\cdot 3^2 + \dots + 9\cdot 3^9 + 10\cdot 3^{10}$.

Subtracting $3S$ from $S$ leads to $-2S = 1 + (2\cdot 3 - 1\cdot 3) + (3\cdot 3^2 - 2\cdot 3^2) + \dots + (10\cdot 3^9 - 9\cdot 3^9) - 10\cdot 3^{10}$, which simplifies to $-2S = 1 + 3 + 3^2 + \dots + 3^9 - 10\cdot 3^{10}$. The sequence $1 + 3 + 3^2 + \dots + 3^9$ is a geometric progression with $10$ terms, and its sum is given by $\frac{1(3^{10}-1)}{3-1} = \frac{3^{10}-1}{2}$. Replacing the geometric series sum within the equation yields $-2S = \frac{3^{10}-1}{2} - 10\cdot 3^{10}$. Rewriting this as $-2S = \frac{3^{10} - 1 - 20\cdot 3^{10}}{2}$ results in $-2S = \frac{-19\cdot 3^{10} - 1}{2}$. Dividing by $-2$ finally produces $S = \frac{19\cdot 3^{10} + 1}{4}$.

Multiplying the recurrence relation $a_{n+2} = \frac{2}{a_{n+1}} + a_n$ by $a_{n+1}$ yields $a_{n+1} a_{n+2} = 2 + a_n a_{n+1}$. This relationship demonstrates that the product of consecutive terms increases by 2 at each step. By defining $x_n = a_n a_{n+1}$, one obtains an arithmetic progression where $x_{n+1} = x_n + 2$, starting with $x_1 = a_1 a_2 = 2$. Each factor in the given product, $\frac{a_k + \frac{1}{a_{k+1}}}{a_{k+2}}$, simplifies to $\frac{a_k a_{k+1} + 1}{a_{k+1} a_{k+2}}$, which is equivalent to $\frac{x_k + 1}{x_{k+1}}$.

Substituting these values generates the product $\frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot s \frac{61}{62}$. Representing the numerator $3 \cdot 5 \cdot \dots \cdot 61$ as $\frac{61!}{2^{30} \cdot 30!}$ and the denominator $4 \cdot 6 \cdot \dots \cdot 62$ as $2^{30} \cdot 31!$ allows the entire expression to be written as $\frac{61!}{2^{60} \cdot 30! \cdot 31!}$. This simplifies to $2^{-60} \binom{61}{31}$, identifying the value of $\alpha$ as $-60$.

The series is a classic example of a telescoping sum, where each term can be simplified using partial fraction decomposition. The general term of this series can be represented as $\frac{1}{n(n+1)(n+2)}$, which is equivalent to $\frac{1}{2} \left( \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right)$. By writing out the summation from $n=2$ to $n=100$, each intermediate term cancels out with the subsequent term, leaving only the first portion of the initial term and the final portion of the last term. This results in the simplified expression $\frac{1}{2} \left( \frac{1}{2 \times 3} - \frac{1}{101 \times 102} \right)$.

Evaluating the expression inside the parentheses gives $\frac{1}{2} \left( \frac{1}{6} - \frac{1}{10302} \right)$. By finding a common denominator, this becomes $\frac{1}{2} \left( \frac{1717 - 1}{10302} \right)$, which simplifies to $\frac{1}{2} \left( \frac{1716}{10302} \right)$, or $\frac{858}{10302}$. Reducing this fraction by dividing both the numerator and the denominator by 6 results in $\frac{143}{1717}$, which can be rewritten as $\frac{143}{17 \times 101}$.

Given that the total sum is equal to $\frac{k}{101}$, equating $\frac{k}{101} = \frac{143}{17 \times 101}$ allows for the direct determination of $k = \frac{143}{17}$. Calculating $34 k$ then becomes a straightforward operation of multiplying $34$ by $\frac{143}{17}$, which simplifies to $2 \times 143$, resulting in 286.

The $k$-th term of a binomial expansion $(a+b)^n$ from the beginning is given by $T_k = \binom{n}{k-1} a^{n-k+1} b^{k-1}$. The $k$-th term from the end is equivalent to the $k$-th term from the beginning when the positions of the terms $a$ and $b$ are swapped, effectively calculating the expansion of $(b+a)^n$. For the expansion $(\frac{4x}{5} - \frac{5}{2x})^{2022}$ with $n=2022$ and $k=1011$, the term from the beginning is $T_{1011} = \binom{2022}{1010} (\frac{4x}{5})^{1012} (-\frac{5}{2x})^{1010}$, while the term from the end is $T'_{1011} = \binom{2022}{1010} (-\frac{5}{2x})^{1012} (\frac{4x}{5})^{1010}$.

Since the term from the end is $2^{10}$ times the term from the beginning, the relationship is expressed as $\binom{2022}{1010} (-\frac{5}{2x})^{1012} (\frac{4x}{5})^{1010} = 2^{10} \cdot \binom{2022}{1010} (\frac{4x}{5})^{1012} (-\frac{5}{2x})^{1010}$. Dividing both sides by the common factor $\binom{2022}{1010} (\frac{4x}{5})^{1010} (-\frac{5}{2x})^{1010}$ simplifies the equation to $(-\frac{5}{2x})^2 = 2^{10} (\frac{4x}{5})^2$. This yields $\frac{25}{4x^2} = 1024 \cdot \frac{16x^2}{25}$, which leads to $x^4 = \frac{5^4}{2^{16}}$. Taking the fourth root of both sides results in $|x| = \frac{5}{16}$.

When the cardinality of the set of pairwise sums $A+A$ equals $2n-1$ for a set of $n$ integers, the elements of $A$ are guaranteed to form an arithmetic progression. In this problem, the set $A$ consists of 20 elements, and the number of pairwise sums is 39, which confirms the set must follow an arithmetic structure.

The sequence starts with 1 and ends with 77 as the 20th term. The common difference $d$ is found using the formula $77 = 1 + 19d$, which leads to $76 = 19d$ and $d = 4$. The intermediate terms $a_1$ through $a_{18}$ form an arithmetic sequence starting at 5 and ending at 73. Calculating their \sum using the formula for the \sum of an arithmetic sequence $S = \frac{n}{2}(first + last)$ gives $S = \frac{18}{2}(5 + 73)$, which simplifies to $9 \times 78 = 702$.

The first term $\frac{6}{3^{12}}$ can be decomposed into $\frac{1}{3^{12}} + \frac{5}{3^{12}}$, allowing the entire expression to be rewritten as $\frac{1}{3^{12}} + 5 \left( \frac{2^0}{3^{12}} + \frac{2^1}{3^{11}} + \frac{2^2}{3^{10}} + \dots + \frac{2^{11}}{3} \right)$. Within the parentheses, each term follows the pattern $\frac{2^k}{3^{12-k}}$, which is equivalent to $\frac{2^k \cdot 3^k}{3^{12}} = \frac{6^k}{3^{12}}$ for values of $k$ ranging from $0$ to $11$. This transforms the expression into $\frac{1}{3^{12}} + \frac{5}{3^{12}} \sum_{k=0}^{11} 6^k$.

Applying the geometric series formula $\sum_{k=0}^{n-1} r^k = \frac{r^n - 1}{r - 1}$ with $r=6$ and $n=12$, the summation $\sum_{k=0}^{11} 6^k$ becomes $\frac{6^{12}-1}{6-1}$, which is $\frac{6^{12}-1}{5}$. Substituting this into the main expression yields $\frac{1}{3^{12}} + \frac{5}{3^{12}} \cdot \left( \frac{6^{12}-1}{5} \right)$. Canceling the factor of 5 simplifies the expression to $\frac{1}{3^{12}} + \frac{6^{12}-1}{3^{12}}$, resulting in $\frac{6^{12}}{3^{12}}$. Since $\frac{6^{12}}{3^{12}} = \left(\frac{6}{3}\right)^{12} = 2^{12}$, we have $2^{12} = 2^n \cdot m$, which implies $n = 12$ and $m = 1$. The final product $m \cdot n$ is therefore $12$.