JEE MAIN - Mathematics (2012 - No. 18)
If $$n$$ is a positive integer, then $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ is :
an irrational number
an odd positive integer
an even positive integer
a rational number other than positive integers
Explanation
Let $${\left( {a + x} \right)^n}$$ = Odd trems(A) + Even terms(B)
So $${\left( {a - x} \right)^n}$$ = Odd terms(A) - Even terms(B)
$$\therefore$$ $${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$$
= (A + B) - (A - B)
= 2B
= 2[even terms]
= 2[ T2 + T4 + T6 + ....... ]
So in case of $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$
= 2[ T2 + T4 + T6 + ....... ]
= 2[ $${}^{2n}{C_1}.{\left( {\sqrt 3 } \right)^{2n - 1}} + {}^{2n}{C_3}.{\left( {\sqrt 3 } \right)^{2n - 3}} + ...$$
Here in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$, 2n - 1 is odd number. So there will be always $${\sqrt 3 }$$ in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$.
So $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ will be always irrational number.
So $${\left( {a - x} \right)^n}$$ = Odd terms(A) - Even terms(B)
$$\therefore$$ $${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$$
= (A + B) - (A - B)
= 2B
= 2[even terms]
= 2[ T2 + T4 + T6 + ....... ]
So in case of $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$
= 2[ T2 + T4 + T6 + ....... ]
= 2[ $${}^{2n}{C_1}.{\left( {\sqrt 3 } \right)^{2n - 1}} + {}^{2n}{C_3}.{\left( {\sqrt 3 } \right)^{2n - 3}} + ...$$
Here in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$, 2n - 1 is odd number. So there will be always $${\sqrt 3 }$$ in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$.
So $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ will be always irrational number.
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