What is the general formula of Binomial Expansion?

Last Updated : 03 Jan, 2024

The Binomial Theorem is used in expanding an expression raised to any finite power. The binomial theorem states that any non-negative power of binomial (x + y)ⁿ can be expanded into a summation of the form $(x+y)^n={^{n}}C_{0}x^{n}+{^{n}}C_{1}x^{n-1}y+{^{n}}C_{2}x^{n-2}y^2+.....+{^{n}}C_{r}x^{n-r}y^r+....+{^{n}}C_{n}y^{n}$ , where n is an integer and each n is a positive integer known as a binomial coefficient. Each term in a binomial expansion is assigned a numerical value known as a coefficient. The number of coefficients in the binomial expansion of (x + y)ⁿ is (n + 1).

General formula of Binomial Expansion

The general form of binomial expansion of (x + y)ⁿ is expressed as a summation function.

$(x+y)^n=\sum_{r=0}^{n} {^{n}}C_{r}x^{n-r}y^r$

where,

n is a positive integer,

x and y are real numbers,

r is an integer such that 0 < r ≤ n.

Derivation

The general formula of binomial expansion can be proved using the principle of mathematical induction.

Let, $S(n) = (x+y)^n=\sum_{r=0}^{n} {^{n}}C_{r}x^{n-r}y^r$

Step 1: Check the given statement S(n) for n = 1.

$(x+y)^1=\sum_{r=0}^{1} {^{1}}C_{r}x^{1-r}y^r\\ ={^{1}}C_{0}x^{1-0}y^0+{^{1}}C_{1}x^{1-1}y^1\\ =(x+y)\\$

So, the result is true for n = 1.

Step 2: Suppose the statement S(n) is true for n = k. So, we get

$(x+y)^k=\sum_{r=0}^{k} {^{k}}C_{r}x^{k-r}y^r\\ ={^{k}}C_{0}x^{k}+{^{k}}C_{1}x^{k-1}y+{^{k}}C_{2}x^{k-2}y^2+.....+{^{k}}C_{r}x^{k-r}y^r+....+{^{k}}C_{k}y^{k}\\ =x^k+{^{k}}C_{1}x^{k-1}y+{^{k}}C_{2}x^{k-2}y^2+....+{^{k}}C_{r}x^{k-r}y^r+....+y^{k}$

Step 3: Now, we have to prove that S(k + 1) is true.

(x + y)^k+1 = (x + y) (x + y)^k

$=(x+y)(x^k+{^{k}}C_{1}x^{k-1}y+{^{k}}C_{2}x^{k-2}y^2+....+{^{k}}C_{r}x^{k-r}y^r+....+y^{k})\\ =x^{k+1}+[1+{^{k}}C_1]x^ky+[{^{k}}C_1+{^{k}}C_2]x^{k-1}y^2+....+[{^{k}}C_{r-1}+{^{k}}C_r]x^{k-r+1}y^r+.....+[{^{k}}C_{k-1}+1]xy^k+y^{k+1}\\ =x^{k+1}+{^{k+1}}C_{1}x^{k}y+{^{k+1}}C_{2}x^{k+1}y^2+.....+{^{k+1}}C_{r}x^{k+1-r}y^r....+{^{k+1}}C_{k}xy^k+y^{k+1}\\ =\sum_{r=0}^{k+1} {^{k+1}}C_{r}x^{k+1-r}y^r$

Thus, the result is true for k+1.

Hence, by mathematical induction the result holds true for all positive integers n.

Sample Problems

Problem 1. Find the binomial expansion of (x + 1)⁴ using the binomial theorem.

Solution:

We have to expand (x + 1)⁴ using the binomial theorem.

Using the binomial expansion, we get

(x+1)⁴ = ⁴C₀ x⁴ + ⁴C₁ x^4-1+ ⁴C₂ x^4-2 + ⁴C₃ x^4-3 + ⁴C₄

=x⁴ + 4x³ + 6x² + 4x + 1

Problem 2. Find the binomial expansion of (3x – 2y)⁴ using the binomial theorem.

Solution:

We have to expand (3x – 2y)⁴ using the binomial theorem.

Using binomial theorem, we have,

(3x – 2y)⁴ = ⁴C₀ (3x)⁴ (2y)⁰ – ⁴C₁ (3x)³ (2y)¹ + ⁴C₂ (3x)² (2y)² – ⁴C₃ (3x)¹ (2y)³ + ⁴C₄ (3x)⁰ (2y)⁴

= 81x⁴ – 4 (27x³) (2y) + 6 (9x²) (4y²) – 4 (3x) (8y³) + 16y⁴

= 81x⁴ – 216x³y + 216x²y² – 96xy³ + 16y⁴

Problem 3. Find the binomial expansion of (4x + 9y)⁵ using the binomial theorem.

Solution:

We have to expand (4x + 9y)⁵ using the binomial theorem.

Using binomial theorem, we have,

(4x + 9y)⁵ = ⁵C₀ (4x)⁵ (9y)⁰ + ⁵C₁ (4x)⁴ (9y)¹+ ⁵C₂ (4x)³ (9y)² + ⁵C₃ (4x)² (9y)³ + ⁵C₄ (4x)¹ (9y)⁴ + ⁵C₅ (4x)⁰ (9y)⁵

= 1024x⁵ + 5 (256x⁴) (9y) + 10 (64x³) (81y²) + 10 (16x²) (729y³) + 5 (4x) (6561y⁴) + 59049 y⁵

= 1024x⁵ +11520x⁴y+51840x³y² +116640x²y³ +131220xy⁴ +59049y⁵

Problem 4. Find the binomial expansion of (1 – 6x)⁶ using the binomial theorem.

Solution:

We have to expand (1 – 6x)⁶ using the binomial theorem.

Using binomial theorem, we have,

(1 – 6x)⁶ = ⁶C₀ (3x)⁰ – ⁶C₁ (6x)¹ + ⁶C₂ (6x)² – ⁶C₃ (6x)³ + ⁶C₄ (6x)⁴ – ⁶C₅ (6x)⁵ + ⁶C₆ (6x)⁶

= 1 – 6 (6x) + 15 (36x²) + 20 (216x³) – 15 (1296x⁵) + 6 (7776x⁵) – (46656x⁶)

= 1 − 36x + 540x² − 4320x³ + 19440x⁴− 46656x⁵ + 46656x⁶.

Problem 5. Find the binomial expansion of (1 – 4x + 2x²)³ using the binomial theorem.

Solution:

We have to expand (1 – 4x + 2x²)³ using the binomial theorem.

Using binomial theorem, we have,

(1 – 4x + 2x²)³ = ³C₀ (1 – 2x)³ + ³C₁ (1 – 2x)² (3x²) + ³C₂ (1 – 2x)(3x²)² + ³C₃ (3x²)³

= (1 – 2x)³ + 9x² (1 – 2x)² + 27x⁴ (1 – 2x) + 27x⁶

= 1 – 8x³ + 12x² – 6x + 9x² (1 + 4x² – 4x) + 27x⁴ – 54x⁵ + 27x⁶

= 1 – 8x³ + 12x² – 6x + 9x² + 36x⁴ – 36x³ + 27x⁴ – 54x⁵ + 27x⁶

= 1 – 6x + 21x² – 44x³ + 63x⁴ – 54x⁵ + 27x⁶

Problem 6. Expand (6x + 4)⁵ using Binomial Theorem.

Solution:

We have to expand (6x + 4)⁵ using the binomial theorem.

Using binomial theorem, we have,

(6x + 4)⁵ = ⁵C₀ (6x)⁵(4⁰) – ⁵C₁ (6x)⁴ (4) + ⁵C₂ (6x)³ (16) – ⁵C₃ (6x)² (64)+ ⁵C₄ (6x)¹ (256) + ⁵C₅ (6x)⁰ (1024)

= 7776 x⁵ + 5(7776x⁵) + 10 (216x³) (16) + 10 (36x²) (64)+ 5 (6x) (256) + 1024

= 7776x⁵ + 25920x⁴ + 34560×3 + 23040×2 + 7680x + 1024.

Problem 7. Expand (y + 7)⁵ using Binomial Theorem.

Solution:

We have to expand (y +7)⁵using the binomial theorem.

Using binomial theorem, we have,

(y + 7)⁵= ⁵C₀ (y)⁵(7⁰) – ⁵C₁ (y)⁴ (7) + ⁵C₂ (y)³ (7)² – ⁵C₃ (y)² (7³)+ ⁵C₄ (y)¹ (7⁴) + 5C5 (y)⁰ (7⁵)

= y⁵ + 5(y⁴)(7) + 10 (y³) (49) + 10 (y²) (343)+ 5 (y) (2401) + 16807

= y⁵ + 35y⁴ + 490y³ + 3430y² + 12005y + 16807