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 + … + ⁿCn αⁿ.
We can conclude that,
(i) (1 + α)ⁿ – 1 = ⁿC₁ α + ⁿC₂ α² + … + ⁿCn 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 + … + ⁿCn 8ⁿ] 9 – 8n – 9
= [1 + 8n + ⁿC₂ 8² + ⁿC₃ 8³ + … + ⁿCn 8ⁿ] 9 – 8n – 9
= 9 + 72n + [ⁿC₂ 8² + ⁿC₃ 8³ + … + ⁿCn 8ⁿ] 9 – 8n – 9
= (72n – 8n) + 8² [ⁿC₂ + ⁿC₃ 8 + … + ⁿCn 8ⁿ⁻²] 9
= 64n + 64 [ⁿC₂ + ⁿC₃ 8 + … + 8ⁿ⁻²] 9
= 64 [n + (ⁿC₂ + ⁿC₃ 8 + … + ⁿCn 8ⁿ⁻²)9]
= 64 x some constant numbers
= Divisible by 64.