Question

State and prove binomial theorem for positive integers.

Answer 1

Statement: “If’n’ is a +ve integer then 

(x + a)ⁿ = ⁿC₀ xⁿ a⁰ + ⁿC₁ x^n-1 a¹ + ⁿC₂ x^n-2 a² + ………….. + ⁿC_n x⁰ aⁿ. 

Proof : (By Mathematical induction): 

Let p(n) : (x + a)ⁿ = ⁿC₀ xⁿ a⁰ + ⁿC₁ x^n-1 a¹ + ⁿC₂ x^n-2a² + ………….. + ⁿC_n x⁰ aⁿ . 

Step1 : P(1) is true 

when n = 1, L.H.S = (x + a)¹ = x + a; RHS = ¹c₀ x¹ a⁰ + ¹_c1 x⁰ a¹ = 1.x.1 + 1.1.a = x+a 

L.H.S. = R.H.S. ⇒ ∴ P(1) is true 

Step 2: Assume that P(m) is true 

i.e., (x + a)^m = ^mc₀ x^m a⁰ + ^mc₁ x^m-1 a₁ + ^mc₂ x^m-2 a² + ………….+ ^mc_mx⁰a^m …….(i) 

Step 3 : P(m + 1)^m+1 is true 

i.e. (x + a)^m+1 = m+¹c₀ x^m+1 a⁰ + ^m+1c₁ x^ma¹ + ^m+1c₂x^m-1a² + …….. + ^m+1c_m+1 x⁰a^m+1 

Multiply both sides of equation (i) by (x+a) 

∴ (x + a)^m (x + a) = (x + a) [^mc₀ x^m a⁰ + ^mc₁ x^m-1 a¹ + ^mc₂ x^m-2 a² + ………….+ ^mc_mx⁰a^m] 

(x + a)^m+1 = x [^mc₀ x^m a⁰ + ^mc₁ x^m-1 a¹ + ^mc₂ x^m-2 a² + ………….+ ^mc_mx⁰a^m] + a[^mc₀ x^ma⁰ + ^mc₁ x^m-1 a¹ + ^mc₂x^m-2 a² + ………….+ ^mc_mx⁰a^m] 

(x + a)^{m + 1} = ^mc₀x^m+1 a⁰ + ^mc₁x^ma¹ + ^mc₂ x^m-1 a² + ……. + ^mc_m x⁰ a^m+1 

= ^mc₀x^m+1a⁰ + (^mc₁ + ^mc₀)x^m a¹ + (^mc₂ + ^mc₁) x^m-1 a² + ………… + (^mc_m + ^mc_m-1) x¹a^m + ^mc_m x⁰a^m+1 

(x + a)^m+1 = ^m+1c₀ x^m+1 a⁰ + ^m+1c¹ x^m a¹ + ^m+1c₂ x^m-1 a² + …….. + ^m+1c_m+1 x⁰ a^m+1 [∴ ^mc_m = ^m+1c₀ ^mc₁ + ^mc₀ = ^m+1c₁ ^mc₂ + ^mc₁ = ^m+1c₂ ^m+1c_m]  

∴ P(m + 1) is true 

Conclusion: P(1) is true, P(m) is true 

⇒ P(m+1) is true 

∴ By principle of mathematical induction the result is true for all natural numbers n. 

Notes: 1. The number of terms in the expansion of (x + a)ⁿ is n + 1.

2. The Gen term of binomial expansion is T_r+1 = ⁿc_r x^n-ra^r .