Key Concept: Binomial Theorem for Differences

The problem requires us to evaluate the nature of the expression ${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$. This form strongly suggests the application of the Binomial Theorem, specifically the expansion of $(x+y)^k - (x-y)^k$.

The Binomial Theorem states that for any positive integer $k$: $(x+y)^k = \binom{k}{0}x^k y^0 + \binom{k}{1}x^{k-1}y^1 + \binom{k}{2}x^{k-2}y^2 + \dots + \binom{k}{k}x^0 y^k$ And $(x-y)^k = \binom{k}{0}x^k y^0 - \binom{k}{1}x^{k-1}y^1 + \binom{k}{2}x^{k-2}y^2 - \dots + (-1)^k \binom{k}{k}x^0 y^k$ When we subtract these two expansions, the terms with even powers of $y$ (which correspond to even values of $r$ in $\binom{k}{r}x^{k-r}y^r$) will cancel out, and the terms with odd powers of $y$ will be added. So, for $k=2n$: $(x+y)^{2n} - (x-y)^{2n} = 2 \left[ \binom{2n}{1}x^{2n-1}y^1 + \binom{2n}{3}x^{2n-3}y^3 + \dots + \binom{2n}{2n-1}x^1 y^{2n-1} \right]$

Step-by-Step Calculation

1. Identify $x$, $y$, and $k$: In our given expression, ${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$, we can identify:

   • $x = \sqrt{3}$
   • $y = 1$
   • $k = 2n$

2. Apply the Binomial Difference Formula: Substitute these values into the derived formula for $(x+y)^k - (x-y)^k$: ${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}} = 2 \left[ \binom{2n}{1}(\sqrt{3})^{2n-1}(1)^1 + \binom{2n}{3}(\sqrt{3})^{2n-3}(1)^3 + \dots + \binom{2n}{2n-1}(\sqrt{3})^1 (1)^{2n-1} \right]$ Explanation: We use this specific formula because it efficiently combines the two binomial expansions and highlights the terms that remain after subtraction. The exponents for $\sqrt{3}$ are $2n-1, 2n-3, \dots, 1$, which are all odd integers. The powers of $1$ are simply $1$.

3. Analyze the terms: Let's look at the general form of a term inside the square brackets: $\binom{2n}{r}(\sqrt{3})^{2n-r}(1)^r$, where $r$ is an odd integer ($1, 3, \dots, 2n-1$). Consider the factor $(\sqrt{3})^{2n-r}$. Since $r$ is odd, $2n-r$ will always be an odd integer. For example:

   • If $2n-r=1$, we have $(\sqrt{3})^1 = \sqrt{3}$.
   • If $2n-r=3$, we have $(\sqrt{3})^3 = (\sqrt{3})^2 \cdot \sqrt{3} = 3\sqrt{3}$.
   • If $2n-r=5$, we have $(\sqrt{3})^5 = (\sqrt{3})^4 \cdot \sqrt{3} = 9\sqrt{3}$. In general, if $m$ is an odd integer, then $(\sqrt{3})^m = 3^{(m-1)/2} \cdot \sqrt{3}$. Explanation: We are examining the nature of each term to determine if it's rational or irrational. The presence of $\sqrt{3}$ raised to an odd power is crucial.

4. Determine the nature of each term and the sum: Each binomial coefficient $\binom{2n}{r}$ is an integer. Each term $\binom{2n}{r}(\sqrt{3})^{2n-r}(1)^r$ will therefore be an integer multiplied by $\sqrt{3}$ (since $(1)^r = 1$). For example, the first term is $2n \cdot \sqrt{3}^{2n-1}$, the second term is $\binom{2n}{3} \cdot \sqrt{3}^{2n-3}$, and so on. Since each term contains a factor of $\sqrt{3}$ (an irrational number) multiplied by a non-zero rational number (an integer), each individual term in the sum is irrational. The sum of irrational numbers of the form $k\sqrt{3}$ (where $k$ is an integer) will also be of the form $K\sqrt{3}$ (where $K$ is an integer). Therefore, the entire sum inside the square brackets will be an irrational number (specifically, a non-zero multiple of $\sqrt{3}$). Multiplying by $2$ (a rational number) does not change its irrational nature, unless the sum was zero, which it is not since $n$ is a positive integer and terms are positive. Explanation: We established that individual terms are irrational. Now we conclude that their sum and the final result will also be irrational, as the structure $A\sqrt{3} + B\sqrt{3} + \dots = (A+B+\dots)\sqrt{3}$ holds, where $A, B, \dots$ are integers derived from the binomial coefficients and powers of 3.

Common Mistakes & Tips

• Forgetting the alternating signs: When subtracting $(x-y)^k$ from $(x+y)^k$, carefully handle the signs. This is why the even-indexed terms cancel and odd-indexed terms double.
• Assuming rationality: Do not quickly assume that products or sums involving irrational numbers will always be irrational. However, in this case, $(\sqrt{3})^{\text{odd integer}}$ always results in a multiple of $\sqrt{3}$.
• Powers of surds: Remember that $(\sqrt{a})^{\text{even integer}}$ is rational, while $(\sqrt{a})^{\text{odd integer}}$ is irrational (if $a$ is not a perfect square).

Summary and Key Takeaway

By applying the Binomial Theorem to the difference of two binomial expansions, we found that all resulting terms contain $\sqrt{3}$ raised to an odd power. This ensures that every term, and consequently their sum, will be an irrational number. Therefore, the given expression is an irrational number.