**Binomial expression **is an algebraic expression with two terms only, e.g. 4x^{2}+9. When such terms are needed to expand to any large power or index say n, then it requires a method to solve it.

Therefore, a theorem called **Binomial Theorem **is introduced which is an efficient way to expand or to multiply a binomial expression. Binomial Theorem is defined as the formula using which any power of a binomial expression can be expanded in the form of a series.

**Binomial theorem**

Binomial Theorem is used to solve binomial expressions in a simple way. It gives an expression to calculate the expansion of (a+b)^{n} for any positive integer n. The Binomial theorem is stated as:

(a + b)

^{n}=^{n}C_{0 }a^{n}+^{n}C_{1 }a^{n-1 }b^{1}+^{n}C_{2 }a^{n-2 }b^{2}+ …. +^{n}C_{r }a^{n-r }b^{r}+ …. +^{n}C_{n }b^{n}

Now, as it is understood that what is a binomial expression and the purpose of the binomial theorem, so try to expand (a+b)^{n} for large values of n (e.g. n = 10. 11, 12,…) using the above statement as:

- (a + b)
^{10}= a^{10}+ 10a^{9}b + 45a^{8}b^{2}+ 120a^{7}b^{3}+ 210a^{6}b^{4}+ 252a^{5}b^{5}+ 210a^{4}b^{6}+120a^{3}b^{7}+ 45a^{2}b^{8}+ 10ab^{9}+ b^{10}- (a + b)
^{11 }= a^{11}+ 11a^{10}b + 55a^{9}b^{2}+ 165a^{8}b^{3}+ 330a^{7}b^{4}+ 462a^{6}b^{5}+ 462a^{5}b^{6}+ 330a^{4}b^{7}+ 165a^{3}b^{8}+ 55a^{2}b^{9}+ 11ab^{10}+ b^{11}- (a + b)
^{12}= a^{12}+ 12a^{11}b + 66a^{10}b^{2}+ 220a^{9}b^{3}+ 495a^{8}b^{4}+ 792a^{7}b^{5}+ 924a^{6}b^{6}+ 792a^{5}b^{7}+ 495a^{4}b^{8}+ 220a^{3}b^{9}+ 66a^{2}b^{10}+ 12ab^{11}+ b^{12}

and so on.

In case of the expansion of smaller powers, it is difficult to calculate the coefficients of the binomial expansion using the same statement. This drawback of the binomial theorem is resolved by Pascal’s Triangle.

The **Pascal Triangle** is an alternative method of the calculation of coefficients that come in binomial expansions, using a diagram rather than algebraic methods.

In the diagram shown below, it is noticed that each number in the triangle is the sum of the two directly above it. This pattern continues indefinitely to obtain coefficients of any index of the binomial expression.

When we observe the pattern of the coefficients of the expansion (a + b)^{n}, The **Pascal’s triangle** for the pattern of the coefficients of the expansion (a + b)^{7 }is shown in the figure below:

Thus, from the above diagram the expansion of small powers of n (e.g. n=0, 1, 2, 3, 4, 5, 6, 7) can be calculated as:

(a + b)^{0}= 1(a + b)^{1 }= a + b(a + b)^{2}= a^{2 }+ 2ab +b^{2}(a + b)^{3}= a^{3}+ 3a^{2}b + 3ab^{2}+ b^{3}- (a + b)
^{4}= a^{4}+ 4a^{3}b + 6a^{2}b^{2}+ 4ab^{3}+ b^{4}- (a + b)
^{5}= a^{5}+ 5a^{4}b + 10a^{3}b^{2}+ 10a^{2}b^{3}+ 5ab^{4}+ b^{5}- (a + b)
^{6}= a^{6}+ 6a^{5}b + 15a^{4}b^{2}+ 20a^{3}b^{3}+ 15a^{2}b^{4}+ 6ab^{5}+ b^{6}- (a + b)
^{7}= a^{7}+ 7a^{6}b + 21a^{5}b^{2}+ 35a^{4}b^{3}+ 35a^{3}b^{4}+ 21a^{2}b^{5}+ 7ab^{6}+ b^{7}

In this way the expansion from (a + b)^{0} to (a + b)^{7} are obtained by the application of Pascal’s Triangle. But finding (a + b)^{15} is really a long process using Pascal’s triangle. So, here is where the binomial theorem comes into the picture.

When the coefficients are observed closely thorem become as follows:

(a+b)

^{1}=^{1}C_{0}a +^{1}C_{1}b(a+b)

^{2}=^{2}C_{0}a^{2}+^{2}C_{1}ab +^{2}C_{2}b^{2}(a+b)

^{3}=^{3}C_{0}a^{3}+^{3}C_{1}a^{2}b +^{3}C_{2}ab^{2}+^{3}C_{3}b^{3}

So from the above pattern, the expansion of (a + b)^{n} becomes as follows:

(a + b)

^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b^{1}+^{n}C_{2}a^{n-2}b^{2}+ …. +^{n}C_{r}a^{n-r}b^{r}+ …. +^{n}C_{n}b^{n}

The above statement is called **Binomial Theorem**.

### Binomial Theorem Formula

The formula obtained by the Binomial Theorem is called the Binomial Theorem Formula, this formula can directly applied to a binomial equation (let it contains terms as x and y) raised to any power n is given as:

Some examples on the above formula are as follows:

- (x+y)
^{2}=x^{2}+2xy+y^{2}- (x+y)
^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}

### Properties of the Binomial Theorem

The Binomial expansion or Binomial theorem is an important part of Arithmetic Mean. It is neither too easy nor too tough because it can be directly applied in a question. Some of the important properties of a Binomial expansion (a+b)^{n} are stated as:

- There are (n+1) terms in an expansion of a binomial expression with index n i.e. one more than the power (index) of the binomial expression. e.g, The number of terms presents in the expansion of (a+b)
^{n}is equal to (n+1). - The expansion of (a+b)
^{n}has first term is equal to a^{n}while the last one is equal to b^{n}. - Such that the sum of the indices of each a and b is equal to n only.
- When a = b or a+b = n only, then
^{n}C_{a }=^{n}C_{b}. - The coefficient of each term equidistant from the beginning and the end are equal. Such coefficients are called as the binomial coefficients and
^{n}C_{r }=^{n}C_{r-1}where r is 0,1,2,…,n.

### Sample Problems on Binomial Theorem

Let’s try some sample problems on the binomial theorem.

**Problem 1: Expand the binomial expression (2x + 3y) ^{2}.**

**Solution:**

(2x)

^{2}+ 2(2x)(3y) + (3y)^{2}= 4x

^{2}+ 12xy + 9y^{2}

**Problem 2: Expand the following (1 – x + x ^{2})^{4}**

**Solution:**

Put 1 – x = y.

Then,

(1 – x + x

^{2})^{4}= (y + x^{2})^{4}=

^{4}C_{0}y^{4}(x^{2})^{0 }+^{4}C_{1}y^{3}(x^{2})^{1 }+^{4}C_{2}y^{2}(x^{2})^{2 }+^{4}C_{3}y(x^{2})^{3 }+^{4}C_{4}(x^{2})^{4}= y

^{4 }+ 4y^{3}x^{2 }+ 6y^{2}x^{4 }+ 4yx^{6 }+ x^{8}= (1 – x)

^{4}+ 4(1 – x)^{3}x^{2}+ 6(1 – x)^{2}x^{4}+ 4(1 – x)x^{6}+ x^{8}= 1 – 4x + 10x

^{2}– 16x^{3}+ 19x^{4}– 16x^{5}+ 10x^{6}– 4x^{7}+ x^{8}

**Problem 3: Find the 4th term from the end in the expansion of ((x ^{3}/2) – (2/x^{2}))^{8}.**

**Solution:**

Since r

^{th}term from the end in the expansion of (a + b)^{n }is(n – r + 2)

^{th}term from the beginning.Therefore, 4

^{th}term from the end is 8 – 4 + 2,i.e., 6th term from the beginning, which is given by

T

_{6}=^{8}C_{5}(x^{3}/2)^{3}(-2/x^{2})^{5}=

^{8}C_{3}(x^{9}/8)(64/x^{10})= 672/x

^{6}

**Problem 4: Find the middle term (terms) in the expansion of ((p/x) + (x/p)) ^{9}**

**Solution:**

Since the power of binomial is odd. Therefore, we have two middle terms

which are 5th and 6th terms. These are given by

T

_{5 }=^{9}C_{4}(p/x)^{5}(x/p)^{4}=

^{9}C_{4}(p/x)= 126(p/x)

T

_{6 }=^{9}C_{5}(p/x)^{4}(x/p)^{5}=

^{9}C_{5}(x/p)= 126(x/p)