First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

an irrational number
B

an odd positive integer
C

an even Positive integer
D

a rational number other than positive integers

Solution

$(\sqrt{3} + 1)^{2 n} =^{2 n} C_{0} (\sqrt{3})^{2 n} +^{2 n} C_{1} (\sqrt{3})^{2 n - 1} +^{2 n} C_{2} (\sqrt{3})^{2 n - 2} + \dots +^{2 n} C_{2 n} (\sqrt{3})^{2 n - 2 n}$
and $(\sqrt{3} - 1)^{2 n} =^{2 n} C_{0} (\sqrt{3})^{2 n} (- 1)^{0} +^{2 n} C_{1} (\sqrt{3})^{2 n - 1} (- 1)^{1}$ $+^{2 n} C_{2} (\sqrt{3})^{2 n - 2} (- 1)^{2} + \dots +^{2 n} C_{2 n} (\sqrt{3})^{2 n - 2 n} (- 1)^{2 n}$
$∴ (\sqrt{3} + 1)^{2 n} - (\sqrt{3} - 1)^{2 n} = 2 [^{2 n} C_{1} (\sqrt{3})^{2 n - 1} +^{2 n} C_{3} (\sqrt{3})^{2 n - 3} +^{2 n} C_{5} (\sqrt{3})^{2 n - 5}$ $+ \dots +^{2 n} C_{2 n - 1} (\sqrt{3})^{2 n - (2 n - 1)}]$
which is most certainly an irrational number because of odd powers of $\sqrt{3}$ in each of the terms.

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