Binomial Theorem and its simple applications - Notes, Formula, Examples, Questions

You have learnt formulas for basic expansions like $(a+b)^2$, $(a-b)^2$,$(a+b)^3$ etc. But what if you need to expand a binomial with a greater power? Expanding binomials with greater power require a mathematical concept called Binomial Theorem. This article is about the mathematics concept of Binomial Theorem Class 11. Binomial Theorem chapter is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, VITEEE, BCECE, and more.

Binomial Theorem is a quick way of expanding a binomial expression with (that are raised to) large powers. This theorem is a really important topic(section) in algebra and has applications in Permutations and Combinations, Probability, Matrices, and Mathematical Induction. If you are preparing for competitive exams for university admission or for jobs then this theorem is really important for you as it is a basic and important section of algebra. In this chapter, you will learn a shortcut that will allow you to find $ (x+y)^n$ without multiplying the binomial by itself $n$ times.

Special cases of the binomial theorem were known since at least the 4th century BC when Greek Mathematician Euclid mentioned the special case of the binomial theorem for exponent $2$. The binomial theorem for cubes was known by the 6th century in India. Isaac Newton is generally credited with the generalized binomial theorem, valid for any rational exponent.

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What is Binomial Theorem?

A binomial is an algebraic expression with two dissimilar terms connected by $+$ or $–$ sign. Let’s look into the following example to understand the difference between monomial, binomial and trinomial.

• $x y^2$ (Monomial term)
• $x-y, y+4$ (Binomial term)
• $x^2+y+1$ (Trinomial term)

First, look into some following identities that you have done earlier

$\begin{aligned}
& (x+y)^0=1 \\
& (x+y)^1=x+y \\
& (x+y)^2=(x+y)(x+y)=x^2+2 x y+y^2 \\
& (x+y)^3=(x+y)\left(x^2+2 x y+y^2\right)=x^3+3 x^2 y+3 x y^2+y^3 \\
& (x+y)^3=(x+y)\left(x^3+3 x^2 y+3 x y^2+y^3\right)=y^4+4 x^3 y+6 x^2 y^2+4 x y^3+y^4
\end{aligned}$

With these smaller powers, it was easy to multiply again and again to get the result. Now, What if you need to find the expansion of $(x+y)^{16}$. Here helps the Binomial Theorem!!!

Now, you will observe that from above expansions,

• The total number of terms in the expansion is one more than the index. For example, in the expansion of $(x+y)^3$, a number of terms is $4$ whereas the index of $(x+y)^2$ is $3$.
• The powers on $x$ begin with $n$ and decrease to $0$ whereas the powers of the second quantity ' $y$ ' begin with $0$ and increase to $n$.
• In each term of the expansion, the sum of the indices of $x$ and $y$ is the same and is equal to the index of $x+y$.
• The combination $\binom{n}{r}$ or ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ occuring in the Binomial theorem is called a Binomial coefficient, where $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$.
• The coefficients are symmetric.

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

Binomial Theorem states that,

If $n$ is any positive integer, then

$ (a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n - 1} b + \binom{n}{2} a^{n - 2} b^2 + \dots + \binom{n}{n - 1} a b^{n - 1} + \binom{n}{n} b^n $

where $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$.

Binomial Theorem Formula

The general formula of binomial theorem to expand binomial expressions with higher power is

$ (a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n - 1} b + \binom{n}{2} a^{n - 2} b^2 + \dots + \binom{n}{n - 1} a b^{n - 1} + \binom{n}{n} b^n $

where $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$.

Binomial Theorem Proof

Statement:

If $n$ is any positive integer, then

$ (a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n - 1} b + \binom{n}{2} a^{n - 2} b^2 + \dots + \binom{n}{n - 1} a b^{n - 1} + \binom{n}{n} b^n $

where $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$.

Proof:

We can prove the binomial theorem using mathematical induction on $n$.
Base Case $(n=1)$ :
For $n=1$, the binomial expansion of $(x+y)^1$ is:

$
(x+y)^1=x+y
$
On the other hand, the binomial formula gives:

$
(x+y)^1=\displaystyle\sum_{k=0}^1\binom{1}{k} x^{1-k} y^k=\binom{1}{0} x^1 y^0+\binom{1}{1} x^0 y^1=x+y
$

So, the base case holds.
Inductive Hypothesis:
Assume that the binomial theorem holds for some integer $n=m$. That is, assume:

$
(x+y)^m=\displaystyle\sum_{k=0}^m\binom{m}{k} x^{m-k} y^k
$

Inductive Step:
We want to show that the binomial theorem holds for $n=m+1$, i.e.,

$
(x+y)^{m+1}=\displaystyle\sum_{k=0}^{m+1}\binom{m+1}{k} x^{(m+1)-k} y^k
$

To do this, consider:

$
(x+y)^{m+1}=(x+y)(x+y)^m
$

Using the inductive hypothesis, we expand $(x+y)^m$ :

$
(x+y)^{m+1}=(x+y) \displaystyle\sum_{k=0}^m\binom{m}{k} x^{m-k} y^k
$

Now distribute $(x+y)$ over the sum:

$
=\displaystyle\sum_{k=0}^m\binom{m}{k} x^{m+1-k} y^k+\displaystyle\sum_{k=0}^m\binom{m}{k} x^{m-k} y^{k+1}
$

We can now combine these two sums into a single sum. Observe that in the first sum, the powers of $x$ start at $m+1$ and decrease, while in the second sum, the powers of $y$ increase starting from 1.
Reindex the second sum by letting $j=k+1$, so that:

$
=\displaystyle\sum_{k=0}^m\binom{m}{k} x^{m+1-k} y^k+\displaystyle\sum_{j=1}^{m+1}\binom{m}{j-1} x^{m+1-j} y^j
$

Now, combine the terms where the powers of $x$ and $y$ match:

$
=x^{m+1}+\displaystyle\sum_{k=1}^m\left(\binom{m}{k}+\binom{m}{k-1}\right) x^{m+1-k} y^k+y^{m+1}
$

Finally, use the identity $\binom{m+1}{k}=\binom{m}{k}+\binom{m}{k-1}$ to rewrite the sum:

$
=\displaystyle\sum_{k=0}^{m+1}\binom{m+1}{k} x^{(m+1)-k} y^k
$
This completes the inductive step.

Thus, by mathematical induction, the binomial theorem holds for all $n \geq 1$.

Properties of Binomial Theorem

Some important properties of binomial theorem are

• The total number of each and every term in the expansion $(x+y)^n$ is $n + 1$ .
• The sum total of the indices of $x$ and $y$ in each term is $n$.
• The expansion shown above is also true when both $x$ and $y$ are complex numbers.
• The coefficient of all the terms is equidistant (equal in distance from each other) from the beginning to the end.
• The values of these binomial coefficients gradually go up to the maximum and progressively lessen.

Terms in Binomial Theorem

General Term in Binomial Expansion

Expanding a binomial with a high exponent such as $(x+2 y)^{16}$ can be a lengthy process. Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term. Note the pattern of coefficients in the expansion of $(x+y)^5$

$
(x+y)^5=x^5+\binom{5}{1} x^4 y+\binom{5}{2} x^3 y^2+\binom{5}{3} x^2 y^3+\binom{5}{4} x y^4+y^5
$

The second term is
$\binom{5}{1} x^4 y$
The third term is
In this way, we can generalize this result.

The $(r+1)$ th term of the binomial expansion of $(x+y)^n$ is:

$
\binom{n}{r} \mathrm{x}^{\mathrm{n}-\mathrm{r}} \mathrm{y}^{\mathrm{r}}
$

Middle term(s) of Binomial Theorem

The middle term in the expansion $(x+y)^n$, depends on the value of ' $n$ '

Case 1: When ' $n$ ' is even
If n is even, and the number of terms in the expansion is $\mathrm{n}+1$, so $\mathrm{n}+1$ is odd number therefore only one middle term is obtained which is

$
\left(\frac{\mathrm{n}}{2}+1\right)_{\text {term. }}^{\text {th }}
$

It is given by

$
\mathrm{T}_{\frac{n}{2}+1}=\binom{n}{\frac{n}{2}} x^{\frac{n}{2}} y^{\frac{n}{2}}
$
Case 2: When ' $n$ ' is odd
In this case, the number of terms in the expansion will be $\mathrm{n}+1$. Since n is odd so, $\mathrm{n}+1$ is even. Therefore, there will be two middle terms in the expansion, namely $\left(\frac{n+1}{2}\right)^{t h}\left(\frac{n+3}{2}\right)^{t h}$ and terms. It is given by

$
T_{\frac{n+1}{2}}=\binom{n}{\frac{n-1}{2}} x^{\frac{n+1}{2}} \cdot y^{\frac{n-1}{2}} \text { and } \quad T_{\frac{n+3}{2}}=\binom{n}{\frac{n+1}{2}} x^{\frac{n-1}{2}} \cdot y^{\frac{n+1}{2}}
$

Independent Term

Independent term is the constant term in the expansion of the binomial expansion. In the expansion of $\left(x+\frac{1}{x}\right)^{2 n}$, where $x \neq 0$, the middle term is $\left(\frac{2 n+1+1}{2}\right)^{t h}$, i.e., $(n+1)^{\mathrm{th}}$ term, as $2 n$ is even.

It is given by ${ }^{2 n} \mathrm{C}_n x^n\left(\frac{1}{x}\right)^n={ }^{2 n} \mathrm{C}_n$ (constant).
This term is called the term independent of $x$ or the constant term.

Numerically Greatest Value

The numerical value of each term of the binomial expansion is determined by the value of the Binomial coefficients. Numerically greatest value is defined as the largest term among the product of the variable coefficients(Binomial coefficients) in the Binomial expansion. In general, Numerically greatest value of the Binomial expansion of $(x+a)^n$ is the $r$th and $(r+1)$th term where $r=\frac{(n+1)}{1+|\frac{x}{a}|}$. It is represented as $T_{r}$ and $T_{r+1}$.

Ratio of Consecutive Terms

The ratio of two consecutive terms in a binomial expansion of the form $(a+b)^n$ is $\frac{n-r+1}{r}$

Application of Binomial Theorem

Finding Remainder Using Binomial Theorem

To find the remainder of an expression using Binomial theorem,

If the number is given in the form of ' $a^n$ ' and which is divided by ' $b$ '. To find the remainder, adjust the power of ' $a$ ' to $a^m$ such that it is very close to ' $b$ ' with a difference of 1 (i.e. $b+1$ or $b-1$).

Also, when number of the type $7 \mathrm{k}-1$ is divided by $7$ , remainder cannot be -1 , as remainder is always positive

So, in such cases, we have $\frac{7 k-1}{7}=\frac{7 k-7+6}{7}=k-1+\frac{6}{7}$

Hence, the reminder is $6(=7-1)$

For example,

If $32^{30}$ is divided by 7 , then the remainder is

We know that $32=2^5$, so, $32^{30}$ can be written as

$\left(2^5\right)^{\underline{30}} =2^{150}=\left(2^3\right)^{50}=8^{50}=(7+1)^{50} $

$ =\left[(7)^{50}+{ }^{50} \mathrm{C}_1(7) \underline{\underline{49}}={ }^{50} \mathrm{C}_2(7) \underline{\underline{48}+\ldots+1]}\right. $

$ =7\left[(7)^{49}+{ }^{50} \mathrm{C}_1(7) \underline{\underline{48}}+{ }^{50} \mathrm{C}_2(7) \underline{\underline{47+}}+1\right.$

$ =7 \mathrm{k}+1$

$ \Rightarrow \text { remainder is } 1 $

Last digits of an expression

To find the last digits of an expansion using Binomial theorem,

If the given expression is $a^n$ then write the expression in the form of $(10 k \pm 1)^m$, where k and m are positive integers

Now take $10$ common for getting last digit, $100$ for getting last two digit, $1000$ for getting last $3$ digits and so on ...

After expanding $\left(10 \mathrm{k} \pm 1{ }^{m}\right.)$ using binomial theorem, it will look like

$(10 \mathrm{k} \pm 1)^{\mathrm{m}}= (10 \mathrm{k})^{\mathrm{m}}+{ }^{\mathrm{m}} \mathrm{C}_1(10 \mathrm{k})^{\mathrm{m}-1}( \pm 1)+{ }^{\mathrm{m}} \mathrm{C}_2(10 \mathrm{k})^{\mathrm{m}-2}( \pm 1)^2 $

$+{ }^{\mathrm{m}} \mathrm{C}_3(10 \mathrm{k})^{\mathrm{m}-2}( \pm 1)^3 \ldots \ldots \ldots+{ }^{\mathrm{m}} \mathrm{C}_{\mathrm{m}-1}(10 \mathrm{k})( \pm 1)^{\mathrm{m}-1}+( \pm 1)^{\mathrm{m}} $

Hence, the number is $10 \alpha+( \pm 1)^m$ (this last part decides the last digit)

The number can also be written as $1008+{ }^m C_{m-1}(10 k)( \pm 1)^{m-1}+( \pm 1)^m$ (last 2 terms decide the last 2 digits) $\alpha, \beta$ are Integers

Binomial Theorem for any Index

Statement: If $n$ is a rational number and $x$ is a real number such that $|\mathrm{x}|<1$, then,

$ (1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots+\frac{n(n-1)(n-2) \ldots \ldots(n-r+1)}{r!} x^r \ldots $

Proof:

Let $f(x)=(1+{{x}})^{{n}}=a_0+a_1 x+a_2 x^2+\ldots+a_1 x^n+\ldots$

$ {{f}(0)}=(1+{{0}} {{{n}}}=1 $

Differentiating (1) w.r.t. $x$ on both sides, we get

$ =a_1+2 a_2 x+3 a_3 x^3+4 a_4 x^3+\ldots+r a_1 x^r-1+\ldots(2) $

Put $x=0$, we get $n=a_1$

Differentiating (2) w.r.t. $\times$ on both sides, we get

$ =2 a_2+6 a_3 x+12 a_4 x^2+\ldots+r(r-1) a_4 x^h-2+\ldots $

Put $x=0$, we get $a_2=[n(n-1)] / 2$ !

Differentiating (3), w.r.t. x on both sides, we get

Put $x=0$, we get $a_3=[n(n-1)(n-2)] / 3$ !

Similarly, we get $a_4=[n(n-1)(n-2)(n-3)] / r!$ and so on

$ \therefore a_n=[n(n-1)(n-2) \ldots(n-r+1)] / r! $

Putting the values of $a_0, a_1, a_2, a_3, \ldots, a_n$ obtained in (1), we get

$ \left(1+x n=1+n x+[\{n(n-1)\} / 2!] x^2+[\{n(n-1)(n-2)\} / 2!] x^3+\ldots+[\{n(n-1)(n-2) \ldots(n-r+\right. $

1) $\} / r!] x^{4}+\ldots$

Hence proved the Binomial theorem of any index.

Real-world Applications of Binomial Theorem

The application of binomial theorem in real life are

1. The binomial theorem is used heavily in Statistical and Probability Analyses. It is so much useful as our economy depends on Statistical and Probability Analyses.

2. In higher mathematics and calculation, the Binomial Theorem is used in finding roots of equations in higher powers. Also, it is used in proving many important equations in physics and mathematics.

3. In Weather Forecast Services,

4. Ranking up candidates

5. Architecture, estimating cost in engineering projects.

List of Topics According to NCERT/JEE MAIN

How To Study Binomial Theorem

First, you need to understand Binomial Theorem. You should be able to find the general term, greatest term, of any binomial equation. After getting a strong understanding of the theorem, you can jump on the application part.

If you are preparing for competitive exams then solve as many problems as you can. Do not jump on the solution right away. Remember if your basics are clear you should be able to solve any question on this topic.

NCERT Notes Subject wise link

• NCERT Notes for class 12 Physics.
• NCERT Notes for class 12 Chemistry.
• NCERT Notes for class 12 Mathematics.
• NCERT Notes for class 12 Biology.

Important Books for Binomial Theorem

Start from NCERT book, the illustration is simple and lucid. You should be able to understand most of the things. Solve all problems (including miscellaneous problem) of NCERT. If you do this, your basic level of preparation will be completed.

Then you can refer to the book Cengage Mathematics Algebra. The binomial theorem is explained very well in this book and there are lots of questions with crystal clear concepts. You can also refer to the book Arihant Algebra by SK Goyal or RD Sharma. But again the choice of reference book depends on person to person, find the book that best suits you the best depending on how well you are clear with the concepts and the difficulty of the questions you require.

NCERT Solutions Subject wise link

• NCERT solutions for class 12 Physics.
• NCERT solutions for class 12 Chemistry.
• NCERT solutions for class 12 Mathematics.
• NCERT solutions for class 12 Biology.