Question

If (5 + 2$\sqrt{6}$)ⁿ = I + f ; n , I Î N; and 0 £ f < 1, then I equals-

• A. $\frac{1}{f}$–f
• B. $\frac{1}{1 + f}$–f
• C. $\frac{1}{1 + f}$+f
• D. $\frac{1}{1 - f}$–f

Answer 1

Answer: (D)

$\frac{1}{1 - f}$–f

Answer 2

Let f¢= (5 –2$\sqrt{6}$)ⁿ

So, I + f + f¢ = (5 + 2$\sqrt{6}$)ⁿ + (5 –2$\sqrt{6}$)ⁿ = an integer

̃ f + f¢ = 1 …(1)

Now, (I + f) × f¢ = (5 + 2$\sqrt{6}$)ⁿ (5 –2$\sqrt{6}$)ⁿ = 1

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