If n is a positive integer,then (sqrt{3} + 1) | vedclass

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(A) Let $x = (\sqrt{3} + 1)^{2n}$ and $y = (\sqrt{3} - 1)^{2n}$.Using the binomial expansion,$(\sqrt{3} + 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sq

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an irrational number

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(A) Let $x = (\sqrt{3} + 1)^{2n}$ and $y = (\sqrt{3} - 1)^{2n}$.Using the binomial expansion,$(\sqrt{3} + 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^{2n-k} (1)^k$.Similarly,$(\sqrt{3} - 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^{2n-k} (-1)^k$.Subtracting the two expressions:$(\sqrt{3} + 1)^{2n} - (\sqrt{3} - 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^{2n-k} [1 - (-1)^k]$.The term $[1 - (-1)^k]$ is $0$ if $k$ is even and $2$ if $k$ is odd.Thus,the expression becomes $2 \sum_{k 	ext{ is odd}} \binom{2n}{k} (\sqrt{3})^{2n-k}$.Since $k$ is odd,$2n-k$ is odd. Let $2n-k = 2m+1$. Then $(\sqrt{3})^{2n-k} = 3^m \sqrt{3}$.However,we can simplify this by noting that $(\sqrt{3} + 1)^2 = 3 + 1 + 2\sqrt{3} = 4 + 2\sqrt{3}$ and $(\sqrt{3} - 1)^2 = 3 + 1 - 2\sqrt{3} = 4 - 2\sqrt{3}$.Let $A = 4 + 2\sqrt{3}$ and $B = 4 - 2\sqrt{3}$. We want $A^n - B^n$.Since $A$ and $B$ are roots of the quadratic equation $x^2 - 8x + 4 = 0$,the expression $A^n - B^n$ is of the form $k\sqrt{3}$ where $k$ is an integer.Wait,let's re-evaluate: $(\sqrt{3}+1)^2 = 4+2\sqrt{3}$. So $(\sqrt{3}+1)^{2n} = (4+2\sqrt{3})^n$.Expanding $(4+2\sqrt{3})^n - (4-2\sqrt{3})^n$ using binomial theorem:$= \sum \binom{n}{k} 4^{n-k} (2\sqrt{3})^k - \sum \binom{n}{k} 4^{n-k} (-2\sqrt{3})^k$$= 2 \sum_{k 	ext{ odd}} \binom{n}{k} 4^{n-k} 2^k (\sqrt{3})^k$.For $k=1$,term is $2 \cdot n \cdot 4^{n-1} \cdot 2 \cdot \sqrt{3} = n \cdot 4^n \sqrt{3}$.Actually,the expression is an irrational number because it contains $\sqrt{3}$.

If $n$ is a positive integer,then $(\sqrt{3} + 1)^{2n} - (\sqrt{3} - 1)^{2n}$ is

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