**Applications of Binomial Theorem**

**(i)** **R-F Factor Relation: **Here, we are going to discuss problems involving (√A + B)ⁿ = I + f, where I and n are positive integers.

0 ≤ f ≤ 1, |A – B²|= k and |√A+B|< 1.

Approach for these types of problems can be learnt from following examples.

**Example**: integral part of (4√3 + 7)ⁿ is (n ϵ N)

**Solution: **“n ϵ N, (7 + 4√3)ⁿ Ï N

Denote (7 + 4√3)ⁿ by I + f

Where, I is an integer and such that 0 < f < 1

∵ 0 < 7 – 4√3 < 1

∴ We can denote (7 + 4√3)ⁿ by G.

Where, G ϵ R such that 0 < G < 1

Now, I + f = (7 + 4√3)ⁿ = 7ⁿ + ⁿC₁ 7ⁿ⁻¹ (4√3) + ⁿC₂ 7ⁿ⁻² (4√3)² + … (i)

G = (7 – 4√3)ⁿ = 7ⁿ – ⁿC₁ 7ⁿ⁻¹ (4√3) + ⁿC₂ 7ⁿ⁻² (4√3)² + … (ii)

To cancel irrational terms we add eqs. (i) and (ii), we get

I + f + G = 2 (7ⁿ + ⁿC₂ 7ⁿ⁻² (48) + ⁿC₄ 7ⁿ⁻⁴ (48)² + …)

= 2k, where k is an integer

∵ I is an integer.

∴ f + G = 2k – I is an integer … (iii)

Now, 0 < f < 1

And 0 < G < 1

⇒ 0 < f + G < 2 … (iv)

From eqⁿ (iii) and (iv), f + G = 1

Now, form eqⁿ (iii) l = 2k – 1

⇒ Integral part of (7 + 4√3)ⁿ.

i.e., l is an odd integer.

**(ii) Divisibility problem: **In the expansion, (1 + α)ⁿ = 1 + ⁿC₁ α + ⁿC₂ α² + … + ⁿC_{n} αⁿ.

We can conclude that,

(i) (1 + α)ⁿ – 1 = ⁿC₁ α + ⁿC₂ α² + … + ⁿC_{n} αⁿ is divisible by α i.e., it is a multiple of α.

**Example:** For all n ϵ R, 9ⁿ⁺¹ – 8n – 9 is divisible by

**Solution: **9ⁿ⁺¹ – 8n – 9 = 9ⁿ x 9 – 8n – 9

= (1 + 8)ⁿ x 9 – 8n – 9

= [ⁿC₀ + ⁿC₁ 8 + ⁿC₂ 8² + ⁿC₃ 8³ + … + ⁿC_{n} 8ⁿ] 9 – 8n – 9

= [1 + 8n + ⁿC₂ 8² + ⁿC₃ 8³ + … + ⁿC_{n} 8ⁿ] 9 – 8n – 9

= 9 + 72n + [ⁿC₂ 8² + ⁿC₃ 8³ + … + ⁿC_{n} 8ⁿ] 9 – 8n – 9

= (72n – 8n) + 8² [ⁿC₂ + ⁿC₃ 8 + … + ⁿC_{n} 8ⁿ⁻²] 9

= 64n + 64 [ⁿC₂ + ⁿC₃ 8 + … + 8ⁿ⁻²] 9

= 64 [n + (ⁿC₂ + ⁿC₃ 8 + … + ⁿC_{n} 8ⁿ⁻²)9]

= 64 x some constant numbers

= Divisible by 64.