If (5 +(2\sqrt{6}))n = I + , where I Ī N, n ĪN and 0 £ <... | MATH - MULTINOMIAL THEOREM, TERMS FREE FROM RADICAL SIGN | SolveeIt

If (5 + $2\sqrt{6}$ )ⁿ = I + , where I Ī N, n ĪN and 0 £ < 1, then I equals –

$\frac{1}{ƒ}$ –

$\frac{1}{1 + ƒ}$ –

$\frac{1}{1–ƒ}$ –

$\frac{1}{1–ƒ}$ +

Answer

$\frac{1}{1–ƒ}$ –

Explanation

We have I + = (5 + $2\sqrt{6}$ )ⁿ

Let G = (5 – $2\sqrt{6}$ )ⁿ

0 < 5 – $2\sqrt{6}$ < 1 Ž 0 < (5 – $2\sqrt{6}$ )ⁿ < 1 Ž 0 < G < 1

I + + G = (5 + $2\sqrt{6}$ )ⁿ + (5 – $2\sqrt{6}$ )ⁿ

= 2 [ⁿC₀ 5ⁿ + ⁿC₂ 5^n–2 ( $2\sqrt{6}$ )² + ….] Ī Z

Ž + G Ī Z (Q I Ī Z)

Now 0 £ < 1, 0 < G < 1 Ž 0 < + G < 2

Ž + G = 1

Now I = (I + ) – = (5 + $2\sqrt{6}$ )ⁿ –

= $\frac{1}{(5–2\sqrt{6})^{n}}$ – = $\frac{1}{G}$ –

= $\frac{1}{1 - ƒ}$ – (Q + G = 1)

Question: If (5 +\(2\sqrt{6}\))<sup>n</sup> = I + , where I Ī N, n ĪN and 0 £  \< 1, then I equals –...