# Binomial Theorem Formula

The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is \((a+b)^{n}=\sum_{r=0}^{n}(^nC_{r})a^{n-r}b^{r}\), where n is a positive integer and a, b are real numbers and 0 < r ≤ n. This formula helps to expand the binomial expressions such as x + a, (2x + 5)^{3}, (x^{ }- (1/x))^{4}, and so on.^{ }

## What Is Binomial Theorem Formula?

The binomial theorem formula helps in the expansion of a binomial raised to a certain power. Let us understand the binomial theorem formula and its application in the following sections.

### Binomial Theorem Formula

The binomial theorem states: If x and y are real numbers, then for all n ∈ N,

(x + y)^{n} = \(^nC_0x^n + ^nC_1x^{n - 1}y + ^nC_2x^{n - 2}y^2 + .....+^nC_rx^{n - r}y^r + ......^nC_nx^0y^n\)

⇒ \((x+ay)^{n}=\sum_{r=0}^{n}(^nC_{r})x^{n-r}y^{r} \)

where, \(^nC_{r}= \dfrac{n!}{ r ! (n-r)!}\)

### Properties of Binomial Theorem

- There are n+1 terms in the expansion of (x+y)
^{n}. - The first and the last terms are x
^{n }and y^{n}respectively. - From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The general term in the expansion of (x + y)
^{n }is the (r +1)^{th}term that can be represented as \(T_{r+1}\),**\(T_{r + 1} = ^nC_rx^{n - r}y^r\).** - The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.
- In the binomial expansion of (x + y)
^{n}, the r^{th}term from the end is (n – r + 2)^{th}term from the beginning. - If n is even, then in (x + y)
^{n}the middle term = (n/2)+1 and if n is odd, then in (x + y)^{n}, the middle terms are (n+1)/2 and (n+3)/2.

## Binomial Theorem Proof

Let x, a, n ∈ N. Let us prove the binomial theorem formula through induction. It is enough to prove for n = 1, n = 2, for n = k ≥ 2, and for n = k+1.

It is obvious that (x +y)^{1} = x +y and

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

= x^{2} + xy +xy + y^{2} (using distributive property)

x^{2} +2xy + y^{2}

Thus the result is true for n = 1 and n = 2. Let k be a positive integer. Let us prove the result is true for k ≥ 2.

Assuming \((x+y)^{n}=\sum_{r=0}^{n}(^nC_{r})x^{n-r}y^{r} \),

\((x+y)^{k}=\sum_{r=0}^{k}(^kC_{r})x^{k-r}y^{r} \),

\((x+y)^{k}= ^nC_0x^k + ^kC_1x^{k - 1}y + ^kC_2x^{k - 2}y^2 + .....+^kC_rx^{k - r}y^r + ......^kC_ky^k\).

\((x+y)^{k}= x^k + ^kC_1x^{k - 1}y + ^kC_2x^{k - 2}y^2 + .....+^kC_rx^{k - r}y^r + .....y^k\)

Thus the result is true for n = k ≥ 2.

Now consider the expansion for k + 1.

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

\(= (x +y)(x^k + ^kC_1x^{k - 1}y + ^kC_2x^{k - 2}y^2 + .....+^kC_rx^{k - r}y^r + ......y^k)\)

\(= x^{k+1} + [1 + ^kC_{1}] x^k y + [^kC_{1 }+^kC_{2}] x ^{k-1} y^2 +........+ [ ^kC_{r-1} + ^kC_{r}] x ^{k-r+1} y ^r + .....\\\\....[^k C _{k-1 }+ 1] xy^k +y^{k+1}\)

\(= x^{k+1 }+ ^{k+1} C _{1}x^{k} y +...... ^{k+1}C_{r} x ^{k+1-r} y^r + ....... ^{k+ 1}C _{k}xy^k +y^{k+1}\)

[\(\because ^nC_r + ^nC_{r-1} = ^{n+1}C_r\)]

Thus the result is true for k+1.

By mathematical induction, this result is true for all positive integers 'n'. Hence the proof.

## Examples Using Binomial Theorem Formula

**Example 1: **Write the expansion of (x+3)^{5 }using the binomial theorem formula.

**Solution:**

\((x+y)^{n}=\sum_{r=0}^{n}(^nC_{r}) x^{n-r}y^{r}\) , where 0 < r ≤ n.

Here a = 3 and n = 5, substituting and expanding, we get:

(x+3)^{5 }= \(^5C_{0} x^{5-0}3^0 + ^5C_{1 }x^{5-1}3^1 + ^5C_{2} x^{5-2}3^2 + ^5C_{3} x^{5-3}3^3 +^5C_{4} x^{5-4}3^4 + ^5C_{5} x^{5-5}3^5\)

= x^{5}^{ }+ 5_{ }x^{4}. 3^{ }+ 10_{ }x^{3 }. 9 + 10_{ }x^{2} . 27^{ }+ 5x .81 +^{ }_{ }3^{5}

= x^{5}^{ }+ 15_{ }x^{4 }+ 90x^{3} + 270_{ }x^{2 }+ ^{ }405 x + 243

**Answer: (x+3) ^{5 }= x^{5}^{ }+ 15_{ }x^{4 }+ 90x^{3} + 270_{ }x^{2 }+ ^{ }405 x + 243**

**Example 2: **Find the 7^{th} term in the expansion of (x + 2)^{10}

**Solution: **

The general term in the expansion of (x+y)^{n }using the binomial theorem formula is

\(T_{r + 1} = ^nC_rx^{n - r}y^r\).

Here r = 6, n =10, a = 2

Thus by substituting, we get

\(T_{7 }= T_{6 + 1} = ^{10}C_6x^{10 -6}2^6\).

T_{7 }= 210 x ^{4 }. 64

= 13440 x^{4}

**Answer: The 7 ^{th} term in the expansion of (x + 2)^{10}13440 x^{4}**

**Example 3: **Find the coefficient of x^{2 }in (x +(1/x))^{8}

**Solution: **

Using the binomial theorem formula in the expansion of (x +1/x)^{8}, we have x^{2 }as the fourth term.

\(^8C_{0} x^8 (1/x)^0 + ^8C_{1} x^7 (1/x)^1 + ^8C_{2} x^6 (1/x)^2 + ^8C_{3} x^5 (1/x)^3\\\\ + ^8C_{4} x^4 (1/x)^4+ ^8C_{5} x^3(1/x)^5 + ^8C_{6} x^2 (1/x)^6 + ^8C_{7} x^1 (1/x)^7 + ^8C_{8} x^0 (1/x)^8\)

The coefficient of the fourth term is \(^8C_{3}\)= 56

**Answer: The coefficient of x ^{2} in the expansion of (1+ (1/x))^{8}= 56**

## FAQs on Binomial Theorem Formula

### What Is the General Form of Binomial Theorem Formula?

the general formula of binomial theorem formula is used in the expansion of binomial expression of the form:

(x + a)^{n} = \(^nC_0x^n + ^nC_1x^{n - 1}a + ^nC_2x^{n - 2}a^2 + .....+^nC_rx^{n - r}a^r + ......^nC_na^n\),

where,

- x, a, n are natural numbers, and,
- 0 < r ≤ n

### Where Is Binomial Theorem Formula Used?

The binomial theorem formula is used to expand the binomial expressions. We apply the formula in finding probability, combinatorics, calculus, and in other important areas of math. For example, (101)^{5 }=(100+1)^{5 }= 100^{5}+ 5 × 100^{4 }+ 10 × 100^{3}+ 10 × 100^{2}+ 5 × 100 + 1 = 10,000,000,000+ 500,000,000 + 10,000,000 + 100,000 + 500 +1 = 10,510,100,501

### What Is n and r in The Binomial Theorem Formula?

In the binomial theorem formula of expansion (x+a)^{n}, we use the combinatorics formula that is denoted as \(^nC_{r}\)_{, }where n is the exponent in the expansion_{ }and_{ }r is the term number that ranges from 0 to n.

### How Do You Find the Number of Terms in the Expansion using the Binomial Theorem Formula?

The number of terms in the binomial expansion of (x+a)^{n }is n+1 terms. i.e. add 1 to the power, the binomial coefficient is raised to. If we expand (2x+5)^{10}, we will have 10 +1 = 11 terms on expansion.