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What is the general formula of Binomial Expansion?

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  • Last Updated : 13 Jun, 2022
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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 x4-1+ 4C2 x4-2 + 4C3 x4-3 + 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)04C1 (3x)3 (2y)1 + 4C2 (3x)2 (2y)24C3 (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)06C1 (6x)1 + 6C2 (6x)26C3 (6x)3 + 6C4 (6x)46C5 (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)5 using the binomial theorem.

Using binomial theorem, we have,

(y + 7)5 = 5C0 (y)5 (70) – 5C1 (y)4 (7) + 5C2 (y)3 (7)25C3 (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


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