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

Binomial Theorem and its simple applications in Mathematics

The Binomial Theorem and its simple applications help in expanding binomial expressions efficiently and finding specific terms or coefficients. This concept is widely used in algebra, probability, and combinatorics, making it important for both academic and competitive exams.

What is the Binomial Theorem?

A binomial is an algebraic expression consisting of two distinct terms connected by $+$ or $–$. Understanding binomials is important before diving into the binomial theorem class 11. Let’s compare monomial, binomial, and trinomial:

Examples:

• $x y^2$ (Monomial)

• $x - y, \ y + 4$ (Binomial)

• $x^2 + y + 1$ (Trinomial)

Basic Binomial Identities

Before learning the binomial theorem formula, consider smaller powers of a binomial:

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

With smaller powers, it is easy to multiply repeatedly. But what if we need the expansion of $(x+y)^{16}$? This is where the binomial theorem helps!

Observations from Binomial Expansions

From the expansions above, we notice:

1. The total number of terms in the expansion of $(x+y)^n$ is $n+1$. For example, $(x+y)^3$ has $4$ terms.

2. The powers of $x$ start from $n$ and decrease to $0$, while powers of $y$ start from $0$ and increase to $n$.

3. In every term, the sum of the indices of $x$ and $y$ equals $n$, the power of the binomial.

4. The coefficients in the expansion are called binomial coefficients, denoted as $\binom{n}{r} = C(n, r) = { }^n C_r = \frac{n!}{r!(n-r)!}$

5. The Binomial coefficient are symmetric, i.e., $\binom{n}{r} = \binom{n}{n-r}$.

These principles form the foundation of the binomial theorem class 11 formulas, useful for binomial theorem examples, proofs, and miscellaneous exercises.

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 the binomial theorem to expand binomial expressions with higher powers 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 the Binomial Theorem

Some important properties of the binomial theorem are

• The total number of each and every term in the expansion $(x+y)^n$ is $n + 1$.
• The 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 the 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 may be interested only in a certain term of a binomial expansion. To find this specific term, we do not need to completely expand the binomial. Now, let us 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 generalise 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, $\mathrm{n}+1$ 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

An 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, the 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

From calculating probabilities, expanding algebraic expressions quickly, to solving problems in combinatorics, the theorem simplifies complex calculations. In this section, you will explore real-life applications of the binomial theorem, its use in class 11 problems, and how it helps in expanding expressions efficiently.

Finding Remainder Using Binomial Theorem

To find the remainder of an expression using the 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 the number of the type $7 \mathrm{k}-1$ is divided by $7$, the remainder cannot be -1, as the 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)^{{30}} =2^{150}=\left(2^3\right)^{50}=8^{50}=(7+1)^{50} $

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

$ =7\left[(7)^{49}+{ }^{50} \mathrm{C}_1(7) {{48}}+{ }^{50} \mathrm{C}_2(7) {{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 the 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 the last digit, $100$ for getting the last two digits, $1000$ for getting the 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 the binomial theorem in real life is:

1. The binomial theorem is used heavily in Statistical and Probability Analyses. It is very 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.

Binomial Theorem and its simple applications in Mathematics: Solved Previous Year Questions

Question 1:

The coefficient of ${x^3}$ in the expansion of $\frac{{{{(1 + 3x)}^2}}}{{1 - 2x}}$ will be:

Solution:

$
\begin{aligned}
& \frac{(1+3 x)^2}{(1-2 x)} \\
& =(1+3 x)^2(1-2 x)^{-1} \\
& =\left(1+6 x+9 x^2\right)\left(1+2 x+4 x^2+8 x^3+\ldots\right)
\end{aligned}
$
$\therefore$ Coefficient of $x^3$ in the given expansion
$
=(8+24+18)=50
$

Hence, the answer is 50.

Question 2:

The coefficient of x in the expansion of ${(1 - ax)^{ - 1}}{(1 - bx)^{ - 1}}{(1 - cx)^{ - 1}}$ is:

Solution:

$
\begin{aligned}
& (1-a x)^{-1}(1-b x)^{-1}(1-c x)^{-1} \\
& \left(1+a x+a^2 x^2+\ldots\right)\left(1+b x+b^2 x^2+\ldots\right)\left(1+c x+c^2 x^2+\ldots\right)
\end{aligned}
$
$\therefore$ Coefficient of $\mathrm{x}=a+b+c$

Hence, the answer is $a+b+c$.

Question 3:

The last digit of $2^{555}+7^{333}$ is:

Solution:

We express the given terms in the form $(10 k \pm 1)^m$.
For $2^{555}$, we write $2=10-8$, so

$
2^{555}=(10-8)^{555}
$

Expanding using binomial theorem:

$
(10-8)^{555}=10^{555}-\binom{555}{1} 10^{554}(8)+\binom{555}{2} 10^{553}\left(8^2\right)-\cdots+(-8)^{555}
$

Since all terms except the last contain a factor of 10 , only the last term determines the last digit:

$
(-8)^{555} \equiv 8^{555} \quad \bmod 10
$

The last digits of powers of 8 cycle as $8,4,2,6$.
Since $555 \bmod 4=3$, the third term in the cycle is 8 .
For $7^{333}$, we write $7=10-3$, so

$
7^{333}=(10-3)^{333}
$

Expanding using binomial theorem:

$
(10-3)^{333}=10^{333}-\binom{333}{1} 10^{332}(3)+\binom{333}{2} 10^{331}\left(3^2\right)-\cdots+(-3)^{333}
$

Only the last term contributes to the last digit:

$
(-3)^{333} \equiv-3^{333} \quad \bmod 10
$

The last digits of powers of 3 cycle as $3,9,7,1$.
Since $333 \bmod 4=1$, the first term in the cycle is 3 .
Thus, the last digit of $(-3)^{333}$ is 7 .
Now, summing the last digits:

$
8+7=15
$

The last digit of 15 is 5.

Hence, the correct answer is 5.

Question 4:

The sum to $(n + 1)$ terms of the following series $\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - \frac{{{C_3}}}{5} + .....$ is:

Solution:

${(1 - x)^n} = {C_0} - {C_1}x + {C_2}{x^2} - {C_3}{x^3} + .......$

Þ $x{(1 - x)^n} = {C_0}x - {C_1}{x^2} + {C_2}{x^3} - {C_3}{x^4} + .....$

$\int_0^1 {x{{(1 - x)}^n}dx = {C_0}\left[ {\frac{{{x^2}}}{2}} \right]_0^1 - {C_1}\left[ {\frac{{{x^3}}}{3}} \right]} _0^1 + {C_2}\left[ {\frac{{{x^4}}}{4}} \right]_0^1 - .......$

The integral on L.H.S. of (i) $=\int_1^0 {(1 - t)\,{t^n}( - dt)} $ by putting $1 - x = t$, Þ $\int_0^1 {({t^n} - {t^{n + 1}})\,dt = \frac{1}{{n + 1}} - \frac{1}{{n + 2}}} $

Whereas the integral on the R.H.S. of (i)

= ${C_0}\left[ {\frac{1}{2}} \right] - {C_1}\left[ {\frac{1}{3}} \right] + \frac{{{C_2}}}{4}.......$ = ${C_0}\left[ {\frac{1}{2}} \right] - {C_1}\left[ {\frac{1}{3}} \right] + \frac{{{C_2}}}{4}.......$$\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - .......$ to $(n + 1)$ terms = $\frac{1}{{n + 1}} - \frac{1}{{n + 2}} = \frac{1}{{(n + 1)\,(n + 2)}}$

Hence, the answer is $ \frac{1}{{(n + 1)\,(n + 2)}}$.

Question 5:

The sum of series $\displaystyle\sum_{\mathrm{r}=0}^{10}{ }^{20} \mathrm{C}_{\mathrm{r}}$ is:

Solution:

$
\begin{aligned}
S & =\sum_{r=0}^{10} 20 r \\
& =\left({ }^{20} C_0+{ }^{20} C_1+\cdots+{ }^{20} C_9\right)+{ }^{20} C_{10} \\
& =S_1+{ }^{20} C_{10}
\end{aligned}
$

Now,
$
\begin{aligned}
& 2 \mathrm{~S}_1+{ }^{20} \mathrm{C}_{10}=2^{20} \\
& S_1+\frac{{ }^{20} C_{10}}{2}=2^{19} \\
& \Rightarrow S_1+{ }^{20} C_{10}=2^{19}+\frac{{ }^{20} C_{10}}{2} \\
& \therefore S=2^{19}+\frac{20}{2} C_{10}
\end{aligned}
$

Hence, the answer is $2^{19}+\frac{{ }^{20} C_{10}}{2}$.

List of Topics related to Binomial Theorem according to NCERT/JEE Mains

Explore the key topics of the binomial theorem class 11 as per the NCERT and JEE Main syllabus. This list includes topics related to the binomial theorem, helping students focus on important areas for exam preparation.

Binomial Theorem in Different Exams

The Binomial Theorem is an important chapter in Class 11 Mathematics and is frequently tested in school exams as well as competitive examinations. It helps in expanding algebraic expressions and solving problems related to coefficients, middle terms, and general terms. The table below presents the exam-wise focus areas, commonly asked topics, and effective preparation strategies to help students study efficiently.

Important Books for Binomial Theorem

Explore the best books for binomial theorem preparation that cover concepts, formulas, examples, and miscellaneous exercises. These resources help students understand the binomial theorem class 11, practice applications, and strengthen problem-solving skills.

NCERT Resources for Binomial Theorem and its simple applications

Access NCERT notes, solutions, and exemplar problems for binomial theorem class 11. These resources provide clear explanations, step-by-step derivations, and examples to help students grasp the binomial theorem formula and its applications.

• NCERT Class 11 Maths Solutions for Chapter 8 Binomial Theorem
• NCERT Class 11 Maths Notes for Chapter 8 Binomial Theorem
• NCERT Exemplar Class 11 Maths Solutions for Chapter 8 Binomial Theorem

NCERT Subjectwise Resources

Find NCERT resources focusing on different subjects for class 11 notes and solutions. These are designed to enhance understanding and provide targeted practice for exams.

Practice Questions based on Binomial Theorem

Practice is key to mastering the binomial theorem in class 11. Solve binomial theorem examples, expansion problems, and real-life applications to strengthen your skills and improve speed and accuracy in exams.

Conclusion

The Binomial Theorem is a powerful tool in algebra that simplifies the expansion of expressions raised to positive integral powers. A clear understanding of its formula, general term, and applications helps students solve problems quickly and accurately. With regular practice and concept-based learning, students can master this chapter and build a strong foundation for advanced mathematical topics and competitive exams.