Key Concept: Binomial Expansion

This problem leverages the Binomial Theorem, specifically the expansion of $(x+y)^n$ and $(x-y)^n$. The Binomial Theorem states that for any non-negative integer $n$: $(x+y)^n = \sum_{k=0}^n {}^n C_k x^{n-k} y^k = {}^n C_0 x^n y^0 + {}^n C_1 x^{n-1} y^1 + {}^n C_2 x^{n-2} y^2 + \dots + {}^n C_n x^0 y^n$ Similarly, for $(x-y)^n$: $(x-y)^n = \sum_{k=0}^n {}^n C_k x^{n-k} (-y)^k = {}^n C_0 x^n y^0 - {}^n C_1 x^{n-1} y^1 + {}^n C_2 x^{n-2} y^2 - \dots + (-1)^n {}^n C_n x^0 y^n$ A useful identity derived from these is the difference: $(x+y)^n - (x-y)^n = 2[{}^n C_1 x^{n-1} y^1 + {}^n C_3 x^{n-3} y^3 + {}^n C_5 x^{n-5} y^5 + \dots]$ This identity includes only the terms with odd powers of $y$.

Problem Transformation

We are asked to find the number of elements $n \in \{1, 2, 3, \dots, 100\}$ such that the inequality $(11)^n > (10)^n + (9)^n$ holds. Assuming (X) n denotes $X^n$, the inequality is: $11^n > 10^n + 9^n$ To make this inequality suitable for binomial expansion, we first rearrange it: $11^n - 9^n > 10^n$ Now, we can apply the binomial expansion identity. Let $x=10$ and $y=1$. Then, $11^n = (10+1)^n$ and $9^n = (10-1)^n$. Substituting these into the rearranged inequality: $(10+1)^n - (10-1)^n > 10^n$ Using the identity for $(x+y)^n - (x-y)^n$ with $x=10$ and $y=1$: $2[{}^n C_1 10^{n-1}(1)^1 + {}^n C_3 10^{n-3}(1)^3 + {}^n C_5 10^{n-5}(1)^5 + \dots] > 10^n$ $2[{}^n C_1 10^{n-1} + {}^n C_3 10^{n-3} + {}^n C_5 10^{n-5} + \dots] > 10^n$ To simplify this expression and make the comparison easier, we divide both sides by $10^n$ (which is always positive for $n \in \{1, \dots, 100\}$, so the inequality direction remains unchanged): $2 \left[ \frac{{}^n C_1 10^{n-1}}{10^n} + \frac{{}^n C_3 10^{n-3}}{10^n} + \frac{{}^n C_5 10^{n-5}}{10^n} + \dots \right] > 1$ $2 \left[ \frac{{}^n C_1}{10} + \frac{{}^n C_3}{10^3} + \frac{{}^n C_5}{10^5} + \dots \right] > 1$ This is the simplified inequality we need to evaluate for different values of $n$.

Step-by-Step Evaluation

Let $S_n = 2 \left[ \frac{{}^n C_1}{10} + \frac{{}^n C_3}{10^3} + \frac{{}^n C_5}{10^5} + \dots \right]$. We need to find $n$ such that $S_n > 1$.

• For $n=1$: The series terminates after the first term since ${}^1 C_k = 0$ for $k > 1$. $S_1 = 2 \left[ \frac{{}^1 C_1}{10} \right] = 2 \left[ \frac{1}{10} \right] = \frac{1}{5} = 0.2$ Since $0.2 \ngtr 1$, the inequality does not hold for $n=1$.

• For $n=2$: The series terminates after the first term since ${}^2 C_k = 0$ for $k > 2$. $S_2 = 2 \left[ \frac{{}^2 C_1}{10} \right] = 2 \left[ \frac{2}{10} \right] = \frac{4}{10} = 0.4$ Since $0.4 \ngtr 1$, the inequality does not hold for $n=2$.

• For $n=3$: The series includes terms for ${}^n C_1$ and ${}^n C_3$. $S_3 = 2 \left[ \frac{{}^3 C_1}{10} + \frac{{}^3 C_3}{10^3} \right] = 2 \left[ \frac{3}{10} + \frac{1}{1000} \right] = 2 [0.3 + 0.001] = 2 \times 0.301 = 0.602$ Since $0.602 \ngtr 1$, the inequality does not hold for $n=3$.

• For $n=4$: The series includes terms for ${}^n C_1$ and ${}^n C_3$. $S_4 = 2 \left[ \frac{{}^4 C_1}{10} + \frac{{}^4 C_3}{10^3} \right] = 2 \left[ \frac{4}{10} + \frac{4}{1000} \right] = 2 [0.4 + 0.004] = 2 \times 0.404 = 0.808$ Since $0.808 \ngtr 1$, the inequality does not hold for $n=4$.

• For $n=5$: The series includes terms for ${}^n C_1$, ${}^n C_3$, and ${}^n C_5$. $S_5 = 2 \left[ \frac{{}^5 C_1}{10} + \frac{{}^5 C_3}{10^3} + \frac{{}^5 C_5}{10^5} \right]$ Recall: ${}^5 C_1 = 5$, ${}^5 C_3 = 10$, ${}^5 C_5 = 1$. $S_5 = 2 \left[ \frac{5}{10} + \frac{10}{1000} + \frac{1}{100000} \right] = 2 [0.5 + 0.01 + 0.00001] = 2 \times 0.51001 = 1.02002$ Since $1.02002 > 1$, the inequality holds for $n=5$.

Analysis for $n > 5$

For $n > 5$, the terms $\frac{{}^n C_1}{10}$, $\frac{{}^n C_3}{10^3}$, etc., will generally increase. Specifically, as $n$ increases, ${}^n C_k$ increases (for $k \le n/2$), and new positive terms will be added to the sum $S_n$. For example, the first term $\frac{{}^n C_1}{10} = \frac{n}{10}$ clearly increases with $n$. Since $S_5 > 1$, and all subsequent terms added to $S_n$ for $n > 5$ are positive, and existing terms like $\frac{n}{10}$ increase, $S_n$ will continue to be greater than 1 for all $n > 5$. Thus, the inequality $11^n > 10^n + 9^n$ holds for all $n \ge 5$.

Determining the Set and Number of Elements

The set of integers $n$ for which the inequality holds is $n \in \{5, 6, 7, \dots, 100\}$. To find the number of elements in this set, we use the formula: Last element - First element + 1. Number of elements $= 100 - 5 + 1 = 96$.

Tips and Common Mistakes

1. Interpretation of Notation: Always clarify ambiguous notation. Here, (X) n was assumed to be $X^n$ based on the topic and context.
2. Careful Binomial Expansion: Ensure all terms are correctly identified and powers are handled accurately. A common mistake is to miss terms or incorrectly apply the signs in $(x-y)^n$.
3. Dividing by Powers: When dividing an inequality by a variable term, ensure that the term's sign is known. If it can be negative, the inequality direction might flip. Here, $10^n$ is always positive.
4. Checking Small Values: For inequalities involving $n$, testing small integer values is a robust way to find the initial values for which the inequality holds. This helps establish a base case for further analysis or proof by induction.
5. Growth Rates: For large $n$, comparing growth rates of functions like $(1.1)^n$ versus $1 + (0.9)^n$ can quickly determine the behavior of the inequality. In this case, $(1.1)^n$ grows exponentially, while $1 + (0.9)^n$ approaches $1$ as $n$ increases, making the inequality hold for larger $n$.

Summary and Key Takeaway

By transforming the initial exponential inequality into a sum of binomial coefficients and evaluating it for small integer values of $n$, we found that the inequality $11^n > 10^n + 9^n$ holds true for all integers $n \ge 5$. Given the domain $n \in \{1, 2, \dots, 100\}$, the set of valid $n$ values is $\{5, 6, \dots, 100\}$. Therefore, the number of elements in this set is 96. This problem demonstrates the practical application of binomial expansions and the importance of checking boundary conditions for inequalities.