Binomial Theorem Class 11th Notes - Free NCERT Class 11 Maths Chapter 8 notes - Download PDF

Students found it relatively easy when they were asked to find the squares or cubes of numbers up to 50. For example, when they have to find (78)⁶ or (127)⁵, they face difficulty due to repeated multiplications. This is where the Binomial theorem makes our life easier. In these NCERT notes on Maths Chapter 7 Class 11, students will study how to expand (x + y)ⁿ or (x - y)ⁿ in an easier way, where n is an integer or a rational number. Binomial theorem class 11 notes also discuss how we get Pascal’s triangle from the expansion of (x + y)ⁿ.

These NCERT notes for class 11 Maths contain many visual graphics to make learning easier for students. Also, the latest CBSE guidelines have been followed by Careers360 experts when they made these notes. Class 11 Math chapter 7 notes help students to find the entire chapter in one place during revision after completing the textbook exercises. Students can also check the NCERT Exemplar Class 11 Maths Chapter 7 Binomial Theorem for extra practice purposes. For full syllabus coverage and solved exercises as well as a downloadable PDF, please visit this link: NCERT.

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Binomial Theorem Class 11 Notes

Careers360 has prepared these Class 11 Binomial Theorem Notes to make your revision smoother and faster.

We know,
$\begin{array}{rl}& (a+b{)}^0=1\\ & (a+b{)}^1=a+b\\ & (a+b{)}^2=a^2+2ab+b^2\\ & (a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3\end{array}$

If we observe each expansion,

• The total number of terms of each expansion is 1 more than the index of $(a+b)$.

• In the expansion, the power of the first quantity is gradually decreasing by 1, and the power of the second quantity is gradually increasing by 1.

• Suppose n is the index of $(a+b)$; then the sum of the indices of a and b of each term of the expansion is n.

Now, arranging the coefficients of expansion with respect to their index.

If we observe the pattern, we can get the coefficient of the next index.

The pattern is given below.

Figure: 1

This diagram is known as Pascal's triangle.

If we apply Pascal's triangle rule, then the coefficients of $(a+b{)}^5$ will be 1 5 10 10 5 1.

Similarly, the coefficient of $(a+b{)}^6$ will be 1 6 15 20 15 6 1.

Now, we can apply the combination formula to find the coefficients.

Binomial coefficient (for a Positive integral index n) ${}^n{C}_r=\frac{n!}{(n-r)!r!}$,
where n and r are positive integers and $0\le r<n$.

Figure 2 can be rewritten as:

Index Coefficients

$\begin{array}{rl}& 0{}^0{C}_0(=1)\\ & 1{}^1{C}_0(=1){}^1{C}_1(=1)\\ & 2{}^2{C}_0(=1){}^2{C}_1(=2){}^2{C}_2(=1)\\ & 3{}^3{C}_0(=1){}^3{C}_1(=3){}^3{C}_3(=3){}^3{C}_3(=1)\\ & 4{}^4{C}_0(=1){}^4{C}_1(=4){}^4{C}_2(=6){}^4{C}_3(=4){}^4{C}_4(=1)\end{array}$

Binomial Theorem

Binomial theorem for any positive integer $n$,
$(a+b{)}^n={}^n{\mathrm{C}}_0{a}^n+{}^n{\mathrm{C}}_1{a}^{n-1}b+{}^n{\mathrm{C}}_2{a}^{n-2}{b}^2+\dots +{}^n{\mathrm{C}}_{n-1}a\cdot b^{n-1}+{}^n{\mathrm{C}}_n{b}^n$

Proof:
The proof is obtained by applying the principle of mathematical induction.
Let the given statement be

$\mathrm{P}(n):(a+b{)}^n={}^n{\mathrm{C}}_0{a}^n+{}^n{\mathrm{C}}_1{a}^{n-1}b+{}^n{\mathrm{C}}_2{a}^{n-2}{b}^2+\dots +{}^n{\mathrm{C}}_{n-1}a\cdot b^{n-1}+{}^n{\mathrm{C}}_n{b}^n$
For $n=1$, we have

$\mathrm{P}(1):(a+b{)}^1={}^1{\mathrm{C}}_0{a}^1+{}^1{\mathrm{C}}_1{b}^1=a+b$
Thus, $\mathrm{P}(1)$ is true.
Suppose $\mathrm{P}(k)$ is true for some positive integer $k$, i.e.

$(a+b{)}^k={}^k{\mathrm{C}}_0{a}^k+{}^k{\mathrm{C}}_1{a}^{k-1}b+{}^k{\mathrm{C}}_2{a}^{k-2}{b}^2+\dots +{}^k{\mathrm{C}}_k{b}^k---(1)$

We shall prove that $\mathrm{P}(k+1)$ is also true, i.e.,

$(a+b{)}^{k+1}={}^{k+1}{\mathrm{C}}_0{a}^{k+1}+{}^{k+1}{\mathrm{C}}_1{a}^{k}b+{}^{k+1}{\mathrm{C}}_2{a}^{k-1}{b}^2+\dots +{}^{k+1}{\mathrm{C}}_{k+1}{b}^{k+1}$

Now, $(a+b{)}^{k+1}=(a+b)(a+b{)}^k$

$\begin{array}{rl}=& (a+b)({}^k{\mathrm{C}}_0{a}^k+{}^k{\mathrm{C}}_1{a}^{k-1}b+{}^k{\mathrm{C}}_2{a}^{k-2}{b}^2+\dots +{}^k{\mathrm{C}}_{k-1}a{b}^{k-1}+{}^k{\mathrm{C}}_k{b}^k)\\ & \text{ from (1)] }\\ =& {}^k{\mathrm{C}}_0{a}^{k+1}+{}^k{\mathrm{C}}_1{a}^{k}b+{}^k{\mathrm{C}}_2{a}^{k-1}{b}^2+\dots +{}^k{\mathrm{C}}_{k-1}{a}^2{b}^{k-1}+{}^k{\mathrm{C}}_{k}a{b}^k+{}^k{\mathrm{C}}_0{a}^{k}b\\ & +{}^k{\mathrm{C}}_1{a}^{k-1}{b}^2+{}^k{\mathrm{C}}_2{a}^{k-2}{b}^3+\dots +{}^k{\mathrm{C}}_{k-1}a{b}^k+{}^k{\mathrm{C}}_k{b}^{k+1}\end{array}$
[by actual multiplication]

$\begin{array}{rl}& ={}^k{\mathrm{C}}_0{a}^{k+1}+({}^k{\mathrm{C}}_1+{}^k{\mathrm{C}}_0)a^{k}b+({}^k{\mathrm{C}}_2+{}^k{\mathrm{C}}_1)a^{k-1}{b}^2+\dots \\ & +({}^k{\mathrm{C}}_k+{}^k{\mathrm{C}}_{k-1})a{b}^k+{}^k{\mathrm{C}}_k{b}^{k+1}[\text{By grouping like terms}]\\ & ={}^{k+1}{\mathrm{C}}_{0}a{a}^{k+1}+{}^{k+1}{\mathrm{C}}_1{a}^{k}b+{}^{k+1}{\mathrm{C}}_2{a}^{k-1}{\mathrm{b}}^2+\dots +{}^{k+1}{\mathrm{C}}_{k}a{b}^k+{}^{k+1}{\mathrm{C}}_{k+1}{b}^{k+1}\end{array}$

(by using ${}^{k+1}{\mathrm{C}}_0=1,{}^k{\mathrm{C}}_r+{}^k{\mathrm{C}}_{r-1}={}^{k+1}{\mathrm{C}}_r$ and ${}^k{\mathrm{C}}_k=1={}^{k+1}{\mathrm{C}}_{k+1}$ )
Thus, it has been proved that $\mathrm{P}(k+1)$ is true whenever $\mathrm{P}(k)$ is true. Therefore, by the principle of mathematical induction, $\mathrm{P}(n)$ is true for every positive integer $n$.

Properties of the Binomial coefficient

1. $n!=n(n-1)(n-2)(n-3)\dots \dots \dots .4\times 3\times 2\times 1$
2. ${}^n{C}_r={}^n{C}_{n-r}$
$\Rightarrow {}^n{C}_r=\frac{n!}{(n-r)!r!}$
and
${}^n{C}_{n-r}=\frac{n!}{(n-r)!(n-n+r)!}$
$\Rightarrow {}^n{C}_{n-r}=\frac{n!}{(n-r)!(r)!}$
Hence, it is proven that both are equal.
3. If ${}^n{C}_x={}^n{C}_y$ only if $x=y$ and $x+y=n$
4. ${}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$
5. If n is a positive integer and $\mathrm{x},\mathrm{y}$ are two complex numbers, then $(x+y{)}^n={}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}y\dots \dots \dots \dots \dots .{}^n{C}_{n-1}x{y}^{n-1}+{}^n{C}_n{y}^n$

Here binomial coefficients are: ${}^n{C}_0,{}^n{C}_1,\dots \dots \dots \dots \dots ..{}^n{C}_{n-1},{}^n{C}_n$

6. Total no. of terms of the given as $(n+1)$ in the expansion
$(x+y{)}^n={}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}y\dots \dots \dots \dots \dots .{}^n{C}_{n-1}x{y}^{n-1}+{}^n{C}_n{y}^n$

To find the middle terms of using the Binomial theorem

The general equation of the binomial is given as:

$\begin{array}{rl}& (x+y{)}^n={}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}y\dots \dots \dots \dots \dots .{}^n{C}_{n-1}x{y}^{n-1}+{}^n{C}_n{y}^n\\ & (x+y{)}^n=T_1+T_2+T_3\dots \dots \dots \dots \dots .T_n+T_{n+1}\end{array}$

There are two cases.
1. If n is odd.

The number of terms of $(a+b{)}^n$ when $n$ is an odd number is $n+1$.
Here, $n+1$ is an even number.
So there will be two middle terms.
That are ${\left(\frac{n+1}{2}\right)}^{\text{th }}$ and ${(\frac{n+1}{2}+1)}^{th}$ terms of the expansion.

2. If n is even

The number of terms of $(a+b{)}^n$ when $n$ is an even number is $n+1$.
Here, $n+1$ is an odd number.
So there will be one middle term.
That is ${\left(\frac{n+1+1}{2}\right)}^{th}={(\frac{n}{2}+1)}^{th}$ term of the expansion.

To find the sum of the coefficient of the binomial terms, we have to put the value of $x$ numerically as one.

For example
Q. If the binomial expression is $(x-y+1{)}^4$, then find the sum of the binomial coefficients is?

Solution: We have to put the value of $x=y=1$ and get the coefficient sum.
$(x-y+1{)}^4$
$\begin{array}{rl}& =(1-1+1{)}^4\\ & =1\end{array}$
So the Sum of the binomial coefficients is 1.

Some Properties of the Binomial coefficients

$(x+y{)}^n={}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}y\dots \dots \dots \dots \dots ..{}^n{C}_{n-1}x{y}^{n-1}+{}^n{C}_n{y}^n$
Put $x=y=1$
$\begin{array}{rl}& \Rightarrow (1+1{)}^n={}^n{C}_0{11}^n+{}^n{C}_{1}1{}^{n-1}1\dots \dots \dots \dots \dots .{}^n{C}_{n-1}{1.1}^{n-1}+{}^n{C}_n{1}^n\\ & \Rightarrow (2{)}^n={}^n{C}_0+{}^n{C}_1\dots \dots \dots \dots \dots .{}^n{C}_{n-1}+{}^n{C}_n\end{array}$

$(x+y{)}^n={}^n{C}_{0}x{x}^n+{}^n{C}_{1}x{x}^{n-1}y\dots \dots \dots \dots \dots .{}^n{C}_{n-1}x{y}^{n-1}+{}^n{C}_n{y}^n$
Put $x=-y=1$
$\begin{array}{rl}& \Rightarrow (1-1{)}^n={}^n{C}_0{1}^n+{}^n{C}_{1}1{}^{n-1}(-1)\dots \dots \dots \dots ..\dots {}^n{C}_{n-1}1.(-1{)}^{n-1}+{}^n{C}_n(-1{)}^n\\ & \Rightarrow 0={}^n{C}_0-{}^n{C}_1\dots \dots \dots \dots \dots ..{}^n{C}_{n-1}(-1{)}^n+{}^n{C}_n(-1{)}^n\end{array}$

Some Particular expansions

$\begin{array}{rl}& (1+x{)}^{-1}=1-x+x^2-x^3\dots \dots ..\\ & (1-x{)}^{-1}=1+x+x^2+x^3\dots \dots ...\\ & (1+x{)}^{-2}=1-2x+3x^2-4x^3.....\\ & (1-x{)}^{-2}=1+2x+3x^2+4x^3.....\end{array}$

Binomial Theorem: Previous Year Question and Answer

Question 1:

Find the term independent of x, where x≠0, in the expansion of ${(\frac{3x^2}{2}-\frac{1}{3x})}^{15}$

Solution:
${(\frac{3x^2}{2}-\frac{1}{3x})}^{15}$............... (Given)
$T_{r+1}{=}^{15}{\text{C}}_r{\left(\frac{3x^2}{2}\right)}^{15-r}{(-\frac{1}{3x})}^r$ ……. (from standard formula of Tr+1)
$\Rightarrow T_{r+1}{=}^{15}{\text{C}}_r{(-1)}^r{3}^{15-2r}{2}^{r-15}{X}^{30-3r}$ …… (i)
Now, for x,
30 – 3r = 0
$\Rightarrow$r = 10
Substituting the value of r in eq (i),

$T_{r+1}{=}^{15}{\text{C}}_{10}{3}^{-5}{2}^{-5}$
$={}^{15}{\text{C}}_{10}{\left(\frac{1}{6}\right)}^5$

Question 2:

Find the value of r, if the coefficients of (2r + 4)^th and (r – 2)^th terms in the expansion of (1 + x)¹⁸ are equal.

Solution:
Given: (1+x)¹⁸
Now, T_(2r+3)+1is the (2r + 4)^th term
Thus, T_(2r+3)+1 = ¹⁸C_2r+3 (x)^2r+3
Now, T_(r-3)+1 is the (r-2)^th term
Thus, T_(r-3)+1 = ¹⁸C_r-3 x^r-3
Now, it is given that the terms are equal in expansion.
Thus,
¹⁸C_2r+3= ¹⁸C_r-3
i.e., 2r + 3 + r – 3 = 18 …… (ⁿC_x = ⁿC_y gives x + y = n)
Thus, 3r = 18
Therefore, r = 6

Question 3:

Find n in the binomial ${(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}})}^n$ if the ratio of 7th term from the beginning to the 7th term from the end is $\frac{1}{6}$

Solution:
Given ${(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}})}^n$
Now, from the beginning,
Seventh term is $T_7=T_{6+1}$
${=}^n{C}_6{\left(\sqrt[3]{2}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^6$...................(i)
& from the end,
The seventh term is the same ,i.e., ${(\frac{1}{\sqrt[3]{3}}+\sqrt[2]{3})}^n$
$T_7{=}^n{C}_6{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n-6}{\left(\sqrt[3]{2}\right)}^6$.............(ii)
Now, it is given that
$\frac{{}^n{C}_6{\left(\sqrt[3]{2}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^6}{{}^n{C}_6{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n-6}{\left(\sqrt[3]{2}\right)}^6}=\frac{1}{6}$
Thus
$\frac{{\left(\sqrt[3]{2}\right)}^{n-12}}{{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n-12}}=\frac{1}{6}$
Thus $(\sqrt[3]{2}\sqrt[3]{3}{)}^{n-12}=6^{-1}$

Thus $6^{\frac{n-12}{3}}=6^{-1}$
$\frac{n-12}{3}=-1$
Thus $n=9$

Importance of NCERT Class 11 Math Chapter 7 Notes

NCERT Class 12 Maths chapter 7 notes are useful in many ways.

• These notes will give conceptual clarity about the Binomial theorem and build a strong base for the Algebra, Calculus, and Probability chapter, which is not only important for board exams in 11 and 12 but also for other competitive exams like JEE Main, NDA, CUET, etc.
• These notes will teach notes how to use the Binomial theorem efficiently to save time by not doing repeated multiplication.
• Students can use these notes for a quick revision whenever they need.
• These notes contain PDF links to other subjects' solutions.
• The latest CBSE 2025-26 guidelines have been followed in these notes.
• Experts who have multiple years of experience in these topics have created these notes so that students can learn these concepts easily.

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