Question
If $$n$$ is a positive integer, then $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ is:
A.
an irrational number
B.
an odd positive integer
C.
an even positive integer
D.
a rational number other than positive integers
Answer :
an irrational number
Solution :
$$\eqalign{
& {\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}} \cr
& = {\left[ {{{\left( {\sqrt 3 + 1} \right)}^2}} \right]^n} - {\left[ {{{\left( {\sqrt 3 - 1} \right)}^2}} \right]^n} \cr
& = {\left( {4 + 2\sqrt 3 } \right)^n} - {\left( {4 - 2\sqrt 3 } \right)^n} \cr
& = {2^n}\left[ {{{\left( {2 + \sqrt 3 } \right)}^n} - {{\left( {2 - \sqrt 3 } \right)}^n}} \right] \cr
& = {2^n} \times 2\left[ {^n{C_1}{2^{n - 1}}\sqrt 3 + {\,^n}{C_3}{{.2}^{n - 3}}3\sqrt 3 + .....} \right] \cr
& = {2^{n + 1}}\sqrt 3 \left[ {^n{C_1}{{.2}^{n - 1}} + {\,^n}{C_3}{2^{n - 3}}.3 + .....} \right] \cr
& = \sqrt 3 \times \,{\text{Some integer}} \cr
& \therefore \,\,{\text{irrational number}} \cr} $$