The expression consisting of two terms is known as binomial expression. For example,

a+b x+y

Binomial expression may be raised to certain powers. For example,

(x+y)^{2 }(a+b)^{5}

## Expansion of Binomial Expression

In order to expand binomial expression, we use repeated multiplication. For example,

(a+b)^{2 }= (a+b)(a+b) = a(a+b) + b(a+b) = a^{2}+ 2ab + b^{2 }(m+n)^{3 }= (m+n)(m+n)^{2 }= (m+n)(m^{2 }+ 2mn + n^{2}) = m(m2 + 2mn + n2) + n(m2 + 2mn + n2) = m^{3}+ 3m^{2}n + 3mn^{2}+ n^{3}

## Binomial Theorem

When power of expression increases, complexity of calculation of binomial expansion increases.To solve this problem, Isaac Newton introduced a theorem known as binomial Theorem.

### Binomial Theorem Statement

For any positive integer n,

(a+x)^{n }= C(n,0)a^{n}+ C(n,1)a^{n-1}x + C(n,2)a^{n-2}x^{2}+ ... + C(n,r)a^{n-1}x^{r}+ ... + C(n,n)x^{n}

### Proof of Binomial Theorem

Binomial theorem can be proved by using Mathematical Induction.

**Principle of Mathematical Induction**

Mathematical induction states that, if P(n) be a statement and if

- P(n) is true for n=1,
- P(n) is true for n=k+1 whenever P(n) is true for n=k.

then P(n) is true for all natural numbers n.

Now, let P(n) be the given statement. Then,

When n=1, LHS = a+x RHS = C(1,0)a+C(1,1)a^{1-1}x = a+x Hence, P(1) is true.

Let us assume that P(n) is true for some value n=k. Then,

(a+x)^{k }= C(k,0)a^{k}+ C(k,1)a^{k-1}x + C(k,2)a^{k-2}x^{2}+ ... + C(k,r)a^{k-1}x^{r}+ ... + C(k,k)x^{k}

Multiplying both sides by (a+x), we get

(a+x)^{k+1 }= (a+x)[C(k,0)a^{k}+ C(k,1)a^{k-1}x + C(k,2)a^{k-2}x^{2}+ ... + C(k,r)a^{k-1}x^{r}+ ... + C(k,k)x^{k}] = C(k,0)a^{k+1}+ C(k,1)a^{k}x + C(k,2)a^{k-1}x^{2 }+^{ ...}+ C(k,k)ax^{k}+ C(k,0)a^{k}x + C(k,1)a^{k-1}x^{2 }+ ... + C(k,k)x^{k+1}= a^{k+1}+ [c(k,1) + c(k,0)]a^{k}x + [c(k,2) + c(k,1)]a^{k-1}x^{2 }+ [c(k,3) + c(k,2)]a^{k-2}x^{3}+ ... + x^{k+1}= C(k+1, 0)a^{k+1}+ C(k+1, 1)a^{k}x + C(k+1, 2)a^{k-1}x^{2}+ ... + C(k+1, r)a^{k+1-r}x^{r}+ C(k+1, k+1)x^{k }

Hence, P(k+1) is true whenever P(k) is true.

So, by principle of mathematical induction P(n) is true for all natural numbers n, i.e.

(a+x)^{n }= C(n,0)a^{n}+ C(n,1)a^{n-1}x + C(n,2)a^{n-2}x^{2}+ ... + C(n,r)a^{n-1}x^{r}+ ... + C(n,n)x^{n}

Similarly,

(a-x)^{n }= C(n,0)a^{n}- C(n,1)a^{n-1}x + C(n,2)a^{n-2}x^{2}- . . . + (-1)^{r}C(n,r)a^{n-1}x^{r}+ ..... + (-1)^{n}C(n,n)x^{n}

### General Term in Binomial Expression

The general term in the expansion of *(a+x) ^{n}* is

*(r+1)*term i.e.

^{th}t_{r+1 }= C(n,r)a^{n-r}x^{r }Thus, First term(r=0), t_{1}= C(n,0)a^{n}Second term(r=1), t_{2}= C(n,1)a^{n-1}x^{1}and so on.

Now, the binomial theorem may be represented using general term as,

**Middle term of Expansion**

In order to find the middle term of the expansion of (a+x)^{n}, we have to consider 2 cases.

**1. When n is even:** When n is even, suppose n = 2m where m = 1, 2, 3, …

Then, number of terms after expansion is 2m+1 which is odd. Thus, it has only one middle term which is (m+1)^{th} term. So,

**2. When n is odd: **When n is odd, suppose n = 2m-1 where m = 1, 2, 3, …

Then, number of terms after expansion is 2m which is even. Thus, it has 2 middle terms which are m^{th} and (m+1)^{th} terms. So,

## Binomial Expansion Examples

1. Expand (a+b)^{5} using binomial theorem.

**Solution:**

Here, the binomial expression is (a+b) and n=5.

So, using binomial theorem we have,

2. Find the middle term of the expansion (a+x)^{10}.

**Solution:**

Since, n=10(even) so the expansion has n+1 = 11 terms. Hence there is only one middle term which is