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

Question

If $n$ is a positive integer, then $(\sqrt3+1)^{2n}-(\sqrt3-1)^{2n}$ is:

Accepted Answer

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.

Solutions · 1