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

Binomial Theorem PYQ — JEE Main 2012 · Shift 0 · Q40 with Solution

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Correct answer: A

Let $a=(\sqrt{3}+1)^{2n},\qquad b=(\sqrt{3}-1)^{2n}.$ We need to determine the nature of $a-b.$
First expand a general term using the binomial theorem: $(\sqrt{3}\pm 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k}(\sqrt{3})^{2n-k}(\pm 1)^k.$ But it is more convenient to observe parity of powers of $\sqrt{3}$ .
Write the difference: $(\sqrt{3}+1)^{2n}-(\sqrt{3}-1)^{2n}.$ Using binomial expansion, terms with even powers of $1$ cancel, and terms with odd powers get doubled. So

=2\sum_{\substack{k=1 \\ k\text{ odd}}}^{2n} \binom{2n}{k}(\sqrt{3})^{2n-k}. $$ 4. Since $k$ is odd, $2n-k$ is odd as well. Therefore every surviving term contains an odd power of $\sqrt{3}$, hence a factor of $\sqrt{3}$. So the whole expression can be written as $$ (\sqrt{3}+1)^{2n}-(\sqrt{3}-1)^{2n}=\sqrt{3}\cdot M, $$ where $M$ is a rational number. In fact, because each term has coefficient $2\binom{2n}{k}3^{(2n-k-1)/2}$, $M$ is actually a positive integer multiple of $2$. 5. Also, the expression is clearly positive because $$\sqrt{3}+1>\sqrt{3}-1>0,$$ so $$ (\sqrt{3}+1)^{2n}>(\sqrt{3}-1)^{2n}. $$ Hence $M>0$. 6. Therefore the given expression is of the form $$\sqrt{3}\times (\text{positive rational number}),$$ which is irrational. 7. Check with a small value, say $n=1$: $$ (\sqrt{3}+1)^2-(\sqrt{3}-1)^2=(4+2\sqrt{3})-(4-2\sqrt{3})=4\sqrt{3}, $$ which is irrational. This supports the result. 8. Option-wise conclusion: - A: an irrational number — **True** - B: an odd positive integer — False - C: an even positive integer — False - D: a rational number other than positive integers — False Hence the correct answer is **A**.

View written solutionFree

Correct answer: A

Let $a=(\sqrt{3}+1)^{2n},\qquad b=(\sqrt{3}-1)^{2n}.$ We need to determine the nature of $a-b.$
First expand a general term using the binomial theorem: $(\sqrt{3}\pm 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k}(\sqrt{3})^{2n-k}(\pm 1)^k.$ But it is more convenient to observe parity of powers of $\sqrt{3}$ .
Write the difference: $(\sqrt{3}+1)^{2n}-(\sqrt{3}-1)^{2n}.$ Using binomial expansion, terms with even powers of $1$ cancel, and terms with odd powers get doubled. So

=2\sum_{\substack{k=1 \\ k\text{ odd}}}^{2n} \binom{2n}{k}(\sqrt{3})^{2n-k}. $$ 4. Since $k$ is odd, $2n-k$ is odd as well. Therefore every surviving term contains an odd power of $\sqrt{3}$, hence a factor of $\sqrt{3}$. So the whole expression can be written as $$ (\sqrt{3}+1)^{2n}-(\sqrt{3}-1)^{2n}=\sqrt{3}\cdot M, $$ where $M$ is a rational number. In fact, because each term has coefficient $2\binom{2n}{k}3^{(2n-k-1)/2}$, $M$ is actually a positive integer multiple of $2$. 5. Also, the expression is clearly positive because $$\sqrt{3}+1>\sqrt{3}-1>0,$$ so $$ (\sqrt{3}+1)^{2n}>(\sqrt{3}-1)^{2n}. $$ Hence $M>0$. 6. Therefore the given expression is of the form $$\sqrt{3}\times (\text{positive rational number}),$$ which is irrational. 7. Check with a small value, say $n=1$: $$ (\sqrt{3}+1)^2-(\sqrt{3}-1)^2=(4+2\sqrt{3})-(4-2\sqrt{3})=4\sqrt{3}, $$ which is irrational. This supports the result. 8. Option-wise conclusion: - A: an irrational number — **True** - B: an odd positive integer — False - C: an even positive integer — False - D: a rational number other than positive integers — False Hence the correct answer is **A**.