Question

If x=(2+√3) ^n, nen and f=x-[x] , then f^2/1-f is

Answer 1

Answer:

$\frac{ {f}^{2} }{1 - f} = 2K - 2$

where

K = [C(n, 0)2ⁿ + C(n, 2)2ⁿ⁻².3 + C(n, 4)2ⁿ⁻⁴.3²·················] ∈ℤ

This value can be calculated for a given value of n.

Step-by-step explanation:

Let x = I + f = (2+√3)ⁿ

where

I is the integral part of x and

f is the fractional part of x

I +f = Σ[C(n, i)2ⁿ⁻ⁱ.(√3)ⁱ ]................(1)

Now let us consider the conjugate of x i.e (2-√3)ⁿ

Since 0 < (2-√3) < 1

⇒ 0 < (2-√3)ⁿ < 1

Let f₁ = (2-√3)ⁿ

⇒ f₁ = Σ[(-1)ⁱC(n, i)2ⁿ⁻ⁱ.(√3)ⁱ] ...........(2)

Adding (1) and (2), we get

I + f + f₁ = 2[C(n, 0)2ⁿ + C(n, 2)2ⁿ⁻².3 + C(n, 4)2ⁿ⁻⁴.3²·················]

It is clear that RHS is free of radicals and hence it is an integer. So, I + f + f₁ is an integer. For a given value of n, the expression on RHS can be calculated. For now let us denote its value by 2K where K is an integer.

I + f + f₁ = 2k ∈ ℤ

f + f₁ = 2K - I ∈ ℤ

Since 0< f<1 and 0<f₁< 1⇒ 0< f + f₁ <2

The only integer that satisfy the above inequality is 1. Hence

f + f₁ = 1 ................................(3)

and

I = 2K - 1 -----------------------(4)

Multiplying I + f and f₁ we get

(I+ f)f₁ = (2+√3)ⁿ (2-√3)ⁿ

(I+ f)(1 - f) = [(2+√3)(2-√3)]ⁿ

(I+ f)(1 - f) = [ 2² - (√3)²]ⁿ

(I+ f)(1 - f) = [ 4 - 3]ⁿ = 1ⁿ = 1

$I + f = \frac{1}{1 - f}$

$I = \frac{1}{1 - f} - f$

$I = \frac{1 - f(1 - f)}{1 - f}$

$I = \frac{(1 - f )+ {f}^{2} }{1 - f}$

$I = 1 + \frac{ {f}^{2} }{1 - f}$

$\frac{ {f}^{2} }{1 - f} = I - 1$

$\frac{ {f}^{2} }{1 - f} = 2K - 1 - 1$

$\frac{ {f}^{2} }{1 - f} = 2K - 2$

where

K = [C(n, 0)2ⁿ + C(n, 2)2ⁿ⁻².3 + C(n, 4)2ⁿ⁻⁴.3²·················]