Question

Let `(5 + 2sqrt(6))^n` = p + f where n∈N and p∈N and 0 < f < 1 then the value of f² – f + pf – p is ______. 

Options

• a natural number

• a negative integer

• a prime number

• an irrational number

Answer 1

Let `(5 + 2sqrt(6))^n` = p + f where n∈N and p∈N and 0 < f < 1 then the value of f² – f + pf – p is a negative integer. 

Explanation:

`(5 + 2sqrt(6))^n` = p + f, n, p∈N, 0 < f < 1

Let f' = `(5 - 2sqrt(6))^n`, 0 < f' < 1

⇒ p + f + f' = `2(5^n + ""^nC_2 5^(n-2)(2sqrt(6))^2 + ...)`

⇒ 2k, k∈N

⇒ p + f + f' = Integer

⇒ f + f' = Integer – p = Integer

But 0 < f + f' < 2

⇒ f + f' = 1

⇒ f' = 1 – f

∴ f² – f + pf – p = (f – 1)(p + f )

= – f'(p + f)

= `-(5 - 2sqrt(6))^n(5 + 2sqrt(6))^n`

= – (25 – 24)ⁿ

= –1

⇒ a negative integer.