Rewritten Solution

1. Key Concepts and Formulas

This problem utilizes the properties of binomial expansion and combinations.

• The general term (or $(r+1)^{th}$ term) in the binomial expansion of $(a+b)^N$ is given by $T_{r+1} = {}^{N}C_r a^{N-r} b^r$.
• For the expansion of $(1+x)^N$, the $(r+1)^{th}$ term is $T_{r+1} = {}^{N}C_r x^r$, and its coefficient is ${}^{N}C_r$.
• Symmetry Property of Combinations: ${}^{N}C_r = {}^{N}C_{N-r}$. This property is useful for simplifying expressions involving binomial coefficients.
• Ratio of Consecutive Combinations: $\frac{{}^{N}C_k}{{}^{N}C_{k-1}} = \frac{N-k+1}{k}$. This identity provides a powerful way to simplify ratios of combination terms.

2. Identifying the Coefficients

The given binomial expansion is $(1+x)^{2n-1}$. Here, the power of the expansion is $N = 2n-1$.

• Coefficient A (30th term): For the $(r+1)^{th}$ term, $r+1 = 30$, which means $r = 29$. Therefore, $A = {}^{2n-1}C_{29}$.

• Coefficient B (12th term): For the $(r+1)^{th}$ term, $r+1 = 12$, which means $r = 11$. Therefore, $B = {}^{2n-1}C_{11}$.

3. Setting up the Equation

We are given the condition $2A = 5B$. Substituting the expressions for $A$ and $B$: $2 \cdot {}^{2n-1}C_{29} = 5 \cdot {}^{2n-1}C_{11}$

4. Step-by-Step Solution

Step 1: Apply the Symmetry Property of Combinations

• Why: To establish a relationship between two combination terms with consecutive lower indices ($k$ and $k-1$), which allows us to use the ratio identity. Directly applying the ratio identity on ${}^{N}C_{29}$ and ${}^{N}C_{11}$ would be complicated as their lower indices are not consecutive.
• We can rewrite ${}^{2n-1}C_{29}$ using the symmetry property ${}^{N}C_r = {}^{N}C_{N-r}$. ${}^{2n-1}C_{29} = {}^{2n-1}C_{(2n-1) - 29} = {}^{2n-1}C_{2n-30}$
• Now, substitute this back into our equation: $2 \cdot {}^{2n-1}C_{2n-30} = 5 \cdot {}^{2n-1}C_{11}$
• Alternative and more direct approach for the ratio identity: Instead of rewriting ${}^{2n-1}C_{29}$ to ${}^{2n-1}C_{2n-30}$, let's consider rewriting it to a form that has a consecutive lower index to ${}^{2n-1}C_{11}$. For this, it is simpler to guess the value of $N=2n-1$ or use the fact that if a solution exists, one of the lower indices might be related to $N$ by the other. Let $N = 2n-1$. The equation is $2 \cdot {}^{N}C_{29} = 5 \cdot {}^{N}C_{11}$. Using the symmetry property ${}^{N}C_r = {}^{N}C_{N-r}$, we can rewrite ${}^{N}C_{29}$ as ${}^{N}C_{N-29}$. So, $2 \cdot {}^{N}C_{N-29} = 5 \cdot {}^{N}C_{11}$. For the ratio identity $\frac{{}^N C_k}{{}^N C_{k-1}}$ to be applicable, we want the lower indices to be consecutive. Let's assume the $k$ in the identity corresponds to $N-29$ and $k-1$ corresponds to $11$. So, $N-29 = k$ and $11 = k-1$. This implies $k=12$. Then $N-29 = 12 \implies N = 41$. If $N=41$, then $2 \cdot {}^{41}C_{12} = 5 \cdot {}^{41}C_{11}$. This implies $\frac{{}^{41}C_{12}}{{}^{41}C_{11}} = \frac{5}{2}$. This is exactly the form needed for the ratio identity.

Step 2: Apply the Ratio of Consecutive Combinations Identity

• Why: This identity allows us to express the ratio of two consecutive combination terms as a simple algebraic expression involving $N$ and $k$, making it easy to solve for $N$.
• Using the identity $\frac{{}^{N}C_k}{{}^{N}C_{k-1}} = \frac{N-k+1}{k}$ with $N=2n-1$, $k=12$ (and $k-1=11$), we have: $\frac{{}^{2n-1}C_{12}}{{}^{2n-1}C_{11}} = \frac{(2n-1)-12+1}{12}$ $\frac{{}^{2n-1}C_{12}}{{}^{2n-1}C_{11}} = \frac{2n-12}{12}$

Step 3: Solve for n

• Why: To find the value of $n$ that satisfies the given condition.
• From Step 1, we established that the problem reduces to: $\frac{{}^{2n-1}C_{12}}{{}^{2n-1}C_{11}} = \frac{5}{2}$
• Now, equate the two expressions for the ratio: $\frac{2n-12}{12} = \frac{5}{2}$
• Multiply both sides by 12: $2n-12 = \frac{5}{2} \cdot 12$ $2n-12 = 5 \cdot 6$ $2n-12 = 30$
• Add 12 to both sides: $2n = 30 + 12$ $2n = 42$
• Divide by 2: $n = \frac{42}{2}$ $n = 21$

5. Tips and Common Mistakes

• Understanding $r$ for the $(r+1)^{th}$ term: Always remember that the $k^{th}$ term corresponds to ${}^{N}C_{k-1}$, not ${}^{N}C_k$. A common mistake is to use the term number directly as $r$.
• Properties of Combinations: Familiarize yourself with the key properties like ${}^{N}C_r = {}^{N}C_{N-r}$ and the ratio identity. These can significantly simplify calculations compared to expanding large factorials.
• Integer Solutions: In binomial expansion problems, $N$ (the power) and $r$ (the lower index of combination) must be non-negative integers. If you get a non-integer value for $n$ or a negative value for $r$, recheck your steps.

6. Summary and Key Takeaway

This problem effectively demonstrates how applying the fundamental properties of binomial coefficients, specifically the symmetry property and the ratio of consecutive combinations, can lead to a quick and elegant solution. Recognizing when to use these properties is crucial for efficiency in competitive exams. The value of $n$ is $\mathbf{21}$.