Rewritten Solution

This problem involves finding coefficients in a binomial expansion and then using those coefficients in a given equation to solve for an unknown integer 'r'. We will systematically break down the problem using the general term formula for binomial expansions.

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1. Understanding the Binomial Expansion and General Term

The binomial theorem provides a formula for expanding expressions of the form $(a+b)^N$. The general term, often denoted as $T_{k+1}$, in the expansion of $(a+b)^N$ is given by: $T_{k+1} = {}^N{C_k} \cdot a^{N-k} \cdot b^k$ where $k$ is the index of the term (starting from $k=0$ for the first term), and ${}^N{C_k} = \frac{N!}{k!(N-k)!}$ is the binomial coefficient.

In our given problem, the expression is ${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}}$. Comparing this with $(a+b)^N$:

• $N = 15$
• $a = 2{x^{{1 \over 5}}}$
• $b = - {1 \over {{x^{{1 \over 5}}}}} = - {x^{ - {1 \over 5}}}$

Now, let's substitute these into the general term formula to find the general term for our expansion: $T_{k+1} = {}^{15}{C_k} \cdot {\left( {2{x^{{1 \over 5}}}} \right)^{15 - k}} \cdot {\left( { - {x^{ - {1 \over 5}}}} \right)^k}$

Explanation: This step is crucial because the general term allows us to determine any specific term's coefficient and the power of $x$ without expanding the entire expression.

Next, we simplify this expression, particularly focusing on collecting the powers of $x$: $T_{k+1} = {}^{15}{C_k} \cdot 2^{15 - k} \cdot \left( {x^{{1 \over 5}}} \right)^{15 - k} \cdot (-1)^k \cdot \left( {x^{ - {1 \over 5}}} \right)^k$ $T_{k+1} = {}^{15}{C_k} \cdot 2^{15 - k} \cdot x^{{{15 - k} \over 5}} \cdot (-1)^k \cdot x^{ - {k \over 5}}$ $T_{k+1} = {}^{15}{C_k} \cdot 2^{15 - k} \cdot (-1)^k \cdot x^{{{15 - k} \over 5} - {k \over 5}}$ $T_{k+1} = {}^{15}{C_k} \cdot 2^{15 - k} \cdot (-1)^k \cdot x^{{{15 - 2k} \over 5}}$

This simplified form of the general term clearly separates the coefficient part (${}^{15}{C_k} \cdot 2^{15 - k} \cdot (-1)^k$) from the variable part ($x^{{{15 - 2k} \over 5}}$ ).

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2. Finding the Coefficient 'm' (for $x^{-1}$)

The problem states that 'm' is the coefficient of the $x^{-1}$ term. To find this, we need to determine the value of $k$ for which the power of $x$ in our general term is $-1$.

Set the exponent of $x$ equal to $-1$: ${{15 - 2k} \over 5} = -1$

Explanation: We equate the exponent of $x$ from the general term to the desired power of $x$ to find which specific term (identified by $k$) has that power.

Now, solve for $k$: $15 - 2k = -5$ $2k = 15 + 5$ $2k = 20$ $k = 10$

So, the $11^{th}$ term (since $k=10$ means $T_{10+1}$) has $x^{-1}$. Now, substitute $k=10$ into the coefficient part of the general term: $m = {}^{15}{C_{10}} \cdot 2^{15 - 10} \cdot (-1)^{10}$

Explanation: Once $k$ is known, we plug it back into the coefficient part of the general term (excluding $x$) to find the actual value of the coefficient. Note the $(-1)^k$ term is crucial for the sign.

$m = {}^{15}{C_{10}} \cdot 2^5 \cdot (1)$ $m = {}^{15}{C_{10}} \cdot 2^5$

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3. Finding the Coefficient 'n' (for $x^{-3}$)

Similarly, 'n' is the coefficient of the $x^{-3}$ term. We repeat the process:

Set the exponent of $x$ equal to $-3$: ${{15 - 2k} \over 5} = -3$

Solve for $k$: $15 - 2k = -15$ $2k = 15 + 15$ $2k = 30$ $k = 15$

Now, substitute $k=15$ into the coefficient part of the general term: $n = {}^{15}{C_{15}} \cdot 2^{15 - 15} \cdot (-1)^{15}$ $n = {}^{15}{C_{15}} \cdot 2^0 \cdot (-1)$

Explanation: Any number raised to the power of 0 is 1 ($2^0=1$). Also, ${}^{N}{C_N} = 1$. The odd power of $(-1)$ results in $-1$. These simplifications are important.

$n = 1 \cdot 1 \cdot (-1)$ $n = -1$

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4. Solving the Given Equation for 'r'

We are given the equation: $m{n^2} = {}^{15}{C_r} \cdot {2^r}$

Now, substitute the values we found for $m$ and $n$: $({}^{15}{C_{10}} \cdot {2^5}) \cdot (-1)^2 = {}^{15}{C_r} \cdot {2^r}$

Explanation: We replace $m$ and $n$ with their calculated values. The $(-1)^2$ term becomes $1$, which simplifies the left side.

$({}^{15}{C_{10}} \cdot {2^5}) \cdot 1 = {}^{15}{C_r} \cdot {2^r}$ ${}^{15}{C_{10}} \cdot {2^5} = {}^{15}{C_r} \cdot {2^r}$

To compare the binomial coefficients, recall the property ${}^N{C_k} = {}^N{C_{N-k}}$. Using this property, we can rewrite ${}^{15}{C_{10}}$ as: ${}^{15}{C_{10}} = {}^{15}{C_{15 - 10}} = {}^{15}{C_5}$

Substitute this back into the equation: ${}^{15}{C_5} \cdot {2^5} = {}^{15}{C_r} \cdot {2^r}$

Explanation: Using the property ${}^N{C_k} = {}^N{C_{N-k}}$ is a common technique in binomial theorem problems to simplify expressions or facilitate comparisons. Here, it makes the comparison more direct.

By comparing both sides of the equation, we can clearly see that: $r = 5$

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5. Key Takeaways and Common Pitfalls

• General Term is Key: Always start by correctly deriving and simplifying the general term ($T_{k+1}$) of the binomial expansion. This is the foundation for finding specific coefficients.
• Exponent Manipulation: Be very careful with algebraic manipulation of exponents. Mistakes here are common.
• Sign Errors: The $(-1)^k$ term is critical. Ensure you correctly evaluate its sign based on whether $k$ is even or odd.
• Binomial Coefficient Properties: Remember useful properties like ${}^N{C_k} = {}^N{C_{N-k}}$, as they can simplify comparisons or calculations significantly.
• Read the Question Carefully: Ensure you are solving for the correct variable (in this case, 'r' from the final equation, not the index 'k' from the general term).

The value of $r$ that satisfies the given condition is $5$.