# What is the binomial theorem and how to expand the algebraic expressions by using this theorem?

In mathematics, a binomial theorem is used for the expansion of the terms like (x + y) n. It is mostly used in statistics and mathematics for probability and statistical analysis and to expand the higher terms.

It is very essential and important because our economy depends on probability and statistical analysis,

Finding the roots of the equations having higher powers, for the higher mathematical calculations, the binomial theorem is used to do this. In this post, we will learn about this theorem and how to expand the higher terms by using it.

## What is the Binomial theorem?

For expanding the higher algebraic expressions like (x + y) n. A question arises in our minds why do we apply the binomial theorem for expanding? The answer is that we can easily expand the lower algebraic expressions such as (x + y)2, and (x + y)3 but it becomes difficult as the power of the exponent’s increases.

So, we have to use this theorem to avoid a large expansion. By using this we can easily expand the higher algebraic expressions like (x + y) n. The terms in the binomial theorem must be the numeric values and are said to be the coefficients of the binomial theorem.

### Statement of Binomial theorem

The binomial theorem states that the expansion of any positive power is possible to expand for (x + y) into an additional form of the coefficients. And the expansion is given as,

(x + y) = nC0 xny0 + nC1 xn-1y1 + nC2 xn-2y2 + nC3 xn-3y3 + … + nCn-1 x1yn-1 + nCn x0yn

In the above equation, n must be greater than 0, nCk is a positive integer known as the coefficient of the binomial.

We can also write the above equation by taking k = 0, 1, 2, 3, …, n – 1, n, and applying summation notation.

$(x+y)^{n}=\sum_{k=0}^{n}{^{n}\textrm{C}}{k}x^{n-k}y^{k}=\sum{k=0}^{n}{^{n}\textrm{C}}_{k}x^{k}y^{n-k}$

To calculate the expansion according to the formula, you can use the binomial expansion calculator.

## What are algebraic expressions?

The expressions that contain one or more variables, numbers, or arithmetic operations by using symbols, powers, or combinations are said to be algebraic expressions. These expressions can be expanded easily for lower powers such as

(x + y)1 = x + y

(x + y)2 = x2 + 2xy + y2

(x + y)3 = x3 + 3xy (x + y) + y3

And so on.

But the expansion for (x + y)5, (x + y)6, (x + y)6, and above become very difficult. So, we use binomial expansion to avoid such a large expansion.

## How to expand the algebraic expressions by using the binomial theorem?

The algebraic expression is of the form of (x + y), for expanding the higher terms, we use the binomial theorem. For using this theorem for expansion, you must have sound knowledge about the sigma notation and combination because all the calculation is based on these terms.

Example 1

Expand the algebraic expression by using binomial theorem when x = 4, y = 2, and n = 6.

Solution

Step 1: Identify the given terms.

x = 4, y = 2, n = 6

Step 2: Put the terms in the general algebraic form such as (x + y)n

(4 + 2)6

Step 3: Identify the values of n.

n = 0, 1, 2, 3, 4, 5

Step 4: Take the general formula of binomial expansion.

$(x+y)^{n}=\sum_{k=0}^{n}{^{n}\textrm{C}}_{k}x^{n-k}y^{k}$

Step 5: Now put the values of n and solve them.

For n = 0

$(4+2)^{0}=\sum_{k=0}^{0}{^{0}\textrm{C}}_{0}4^{0-0}2^{0}$

(4 + 2)0 = 1

For n = 1

$(4+2)^{1}=\sum_{k=0}^{1}{^{1}\textrm{C}}_{0}4^{1-0}2^{0}+{^{1}\textrm{C}}_{1}4^{1-1}2^{1}$

(4 + 2)1 = (1)(4) + (1)(1)(2)

(4 + 2)1 = 4 + 2

(4 + 2)1 = 6

For n = 2

$(4+2)^{2}=\sum_{k=0}^{2}{^{2}\textrm{C}}_{0}4^{2-0}2^{0}+{^{2}\textrm{C}}_{1}4^{2-1}2^{1}+{^{2}\textrm{C}}_{2}4^{2-2}2^{2}$

(4 + 2)2 = (1)(16)(1) + (2)(4)(2) + (1)(1)(4)

(4 + 2)2 = 16 +16 + 4

(4 + 2)2 = 36

For n = 3

$(4+2)^{3}=\sum_{k=0}^{3}{^{3}\textrm{C}}_{0}4^{3-0}2^{0}+{^{3}\textrm{C}}_{1}4^{3-1}2^{1}+{^{3}\textrm{C}}_{2}4^{3-2}2^{2}+{^{3}\textrm{C}}_{3}4^{3-3}2^{3}$

(4 + 2)3 = (1)(43)(1) + (3)(42)(2) + (3)(4)(22) + (1)(1)(23)

(4 + 2)3 = 64 + 6(16) + 12(4) + (8)

(4 + 2)3 = 64 + 96 + 48 + 8

(4 + 2)3 = 216

For n = 4

$(4+2)^{4}=\sum_{k=0}^{4}{^{4}\textrm{C}}_{0}4^{4-0}2^{0}+{^{4}\textrm{C}}_{1}4^{4-1}2^{1}+{^{4}\textrm{C}}_{2}4^{4-2}2^{2}+{^{4}\textrm{C}}_{3}4^{4-3}2^{3}+{^{4}\textrm{C}}_{4}4^{4-4}2^{4}$

(4 + 2)4 = (1)(44)(1) + (4)(43)(2) + (6)(42)(22) + (4)(4)(23) + (1)(1)(24)

(4 + 2)4 = 256 + 8(64) + 6(16)(4) + 16(8) + 16

(4 + 2)4 = 256 + 512 + 384 + 128 + 16

(4 + 2)4 = 1296

For n = 5

$(4+2)^{5}=\sum_{k=0}^{5}{^{5}\textrm{C}}_{0}4^{5-0}2^{0}+{^{5}\textrm{C}}_{1}4^{5-1}2^{1}+{^{5}\textrm{C}}_{2}4^{5-2}2^{2}+{^{5}\textrm{C}}_{3}4^{5-3}2^{3}+{^{5}\textrm{C}}_{4}4^{5-4}2^{4}+{^{5}\textrm{C}}_{5}4^{5-5}2^{5}$

(4 + 2)5 = (1)(45)(1) + (5)(44)(2) + (10)(43)(22) + (10)(42)(23) + (5)(4)(24) + (1)(1)(25)

(4 + 2)5 = 1024 + 10(256) + 10(64)(4) + 10(16)(8) + 20(16) + 32

(4 + 2)5 = 1024 + 2560 + 2560 + 1280 + 320 + 32

(4 + 2)5 = 7776

For n = 6

$(4+2)^{6}=\sum_{k=0}^{6}{^{6}\textrm{C}}_{0}4^{6-0}2^{0}+{^{6}\textrm{C}}_{1}4^{6-1}2^{1}+{^{6}\textrm{C}}_{2}4^{6-2}2^{2}+{^{6}\textrm{C}}_{3}4^{6-3}2^{3}+{^{6}\textrm{C}}_{4}4^{6-4}2^{4}+{^{6}\textrm{C}}_{5}4^{6-5}2^{5}+{^{6}\textrm{C}}_{6}4^{6-6}2^{6}$

(4 + 2)6 = (1)(46)(1) + (6)(45)(2) + (15)(44)(22) + (20)(43)(23) + (15)(42)(24) + (6)(4)(25) + (1)(1)(26)

(4 + 2)6 = 4096 + 12(1024) + 15(256) (4) + 20(64)(8) + 15(16)(16) + 24(32) + 64

(4 + 2)6 = 4096 + 12288 + 15360 + 10240 + 3840 + 768 + 64

(4 + 2)6 = 46656

Step 6: Result.

(4 + 2)6 = 46656

Hence, the expansion of 4 and 2 having power 6 is 46656.

## Summary

The binomial theorem is very useful for the expansion of the higher powers of the algebraic expressions. From the above example, you must be witnessed that the calculations of the binomial expansion are not difficult but a little bit lengthy. You can solve any kind of problem-related to expansions by following the above example.