Binomial Theorem Class 11 Notes CBSE Maths Chapter 7 (Free PDF Download)

• The computations become harder when employing repeated multiplication for higher powers like ${{\left( 93 \right)}^{8}},{{\left( 105 \right)}^{7}},{{\left( 305 \right)}^{4}},...$ and so on.
  

• A theorem known as the binomial theorem was used to solve this problem. 

• It simplifies the expansion of ${{\left( \text{a+b} \right)}^{\text{n}}}$, where $\text{n}$ is an integer or a rational number.

Binomial Theorem

• If \[\text{a, b }\in \text{ R}\]and \[\text{n }\in \text{ N}\] , then

\[{{\left( \text{a+b} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{a}}^{\text{n}}}{{\text{b}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{a}}^{\text{0}}}{{\text{b}}^{\text{n}}}\]

• Remarks :

1. If the binomial's index is $\text{n}$, the expansion will have $\text{n+1}$ terms .

2. The total sum of the indices of $\text{a}$ and $\text{b}$ in each term is always $\text{n}$.

3. In a binomial expansion equidistant from both ends, the coefficients of the terms are equal.

4. \[{{\left( \text{a-b} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{a}}^{\text{n}}}{{\text{b}}^{\text{0}}}-{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}\left( -1 \right){}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{a}}^{\text{0}}}{{\text{b}}^{\text{n}}}\]

Binomial Theorem for Positive Integral Indices

• \[{{\left( \text{a+b} \right)}^{\text{0}}}\text{=1}\], where \[\text{a+b }\ne 0\]

\[{{\left( \text{a+b} \right)}^{\text{1}}}\text{= a+b}\]

\[{{\left( \text{a+b} \right)}^{\text{2}}}\text{= }{{\text{a}}^{\text{2}}}\text{+2ab+}{{\text{b}}^{\text{2}}}\]

\[{{\left( \text{a+b} \right)}^{\text{3}}}\text{= }{{\text{a}}^{\text{3}}}\text{+3}{{\text{a}}^{\text{2}}}\text{b+3a}{{\text{b}}^{\text{2}}}\text{+}{{\text{b}}^{\text{3}}}\]

\[{{\left( \text{a+b} \right)}^{\text{4}}}\text{=}{{\left( \text{a+b} \right)}^{\text{3}}}\text{. }\left( \text{a+b} \right)\text{=}{{\text{a}}^{\text{4}}}\text{+4}{{\text{a}}^{\text{3}}}\text{b+6}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{+4a}{{\text{b}}^{\text{3}}}\text{+}{{\text{b}}^{\text{4}}}\]

• We can see in these expansions that

1. There are one more terms in the expansion than the terms present in the index.

2. In each term, the powers of the first quantity ‘$\text{a}$' decrease by one, while the powers of the second quantity ‘$\text{b}$' increase by one.

3. The index of $\left( \text{a+b} \right)$ is equal to the total of the indices of $\text{a}$ and $\text{b}$ in each term of the expansion.

Pascal’s Triangle:

• The expansion coefficients are organised in a triangle-shaped array with $\left( \text{1} \right)$ at the top vertex and running down the two slanting sides. This arrangements is known as “Pascal's triangle”.

• Pingla also refers to it as Meru Prastara.

• Pascal's triangle can also be used to expand binomials to their higher powers.

• Here, Combination method is used.

\[{}^{\text{n}}{{\text{C}}_{\text{r}}}\text{=}\dfrac{\text{n!}}{\text{r!}\left( \text{n-r} \right)\text{!}}\] , \[\text{0}\le \text{r}\le \text{n}\]

Where, \[\text{n}\]is a non-negative number.

• \[{}^{\text{n}}{{\text{C}}_{\text{0}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{n}}}\text{=1}\]

Binomial Theorem for any Positive Integer\[n\]:

• \[{{\left( \text{a+b} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{a}}^{\text{n}}}{{\text{b}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{a}}^{\text{0}}}{{\text{b}}^{\text{n}}}\]

That is,

\[{{\left( \text{a+b} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{a}}^{\text{n}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{b}}^{\text{n}}}\]

• Remarks:

1. Binomial theorem can also be written as,

\[{{\left( \text{a+b} \right)}^{\text{n}}}=\sum\limits_{\text{k=0}}^{\text{n}}{{}^{\text{n}}{{\text{C}}_{\text{k}}}{{\text{a}}^{\text{n-k}}}{{\text{b}}^{\text{k}}}}\]

Where, \[\sum\limits_{\text{k=0}}^{\text{n}}{{}^{\text{n}}{{\text{C}}_{\text{k}}}{{\text{a}}^{\text{n-k}}}{{\text{b}}^{\text{k}}}}\] represents 

\[{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{a}}^{\text{n}}}{{\text{b}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{a}}^{\text{0}}}{{\text{b}}^{\text{n}}}\]

2. The coefficients \[{}^{\text{n}}{{\text{C}}_{\text{r}}}\]are known as binomial coefficients.

Example: Compute \[{{\left( 99 \right)}^{5}}\]

Ans:

We can consider \[99\] as follows,

\[99=100-1\]

Therefore, 

\[{{\left( 99 \right)}^{5}}={{\left( 100-1 \right)}^{5}}\]

By applying Binomial Theorem for above, we get

${{\left( 99 \right)}^{5}}={}^{5}{{\text{C}}_{0}}{{\left( 100 \right)}^{5}}{{.1}^{0}}-{}^{5}{{\text{C}}_{1}}{{\left( 100 \right)}^{4}}{{.1}^{1}}+{}^{5}{{\text{C}}_{2}}{{\left( 100 \right)}^{3}}{{.1}^{2}}-{}^{5}{{\text{C}}_{3}}{{\left( 100 \right)}^{2}}{{.1}^{3}} +{}^{5}{{\text{C}}_{4}}{{\left( 100 \right)}^{1}}{{.1}^{4}}+{}^{5}{{\text{C}}_{5}}{{\left( 100 \right)}^{0}}{{.1}^{5}}$

$ =100000000005\times 100000000\times 1+10\times 1000000\times 110\times 10000\times 1  +5\times 100\times 11 $

\[=10000000000-500000000+10000000-100000+500-1\]

\[=9509900499\]

Therefore,

\[{{\left( 99 \right)}^{5}}=9509900499\]

General Term and Middle Terms in Expansion of \[{{\left( \text{a+b} \right)}^{\text{n}}}\]:

• \[{{\text{t}}_{\text{r+1}}}\text{= }{{\text{ }}^{\text{n}}}{{\text{C}}_{\text{r}}}{{\text{a}}^{\text{n-r}}}+{{\text{b}}^{\text{r}}}\]

• \[{{\text{t}}_{\text{r+1}}}\] is known as a general term for all \[\text{r}\in \text{N}\] and \[\text{0}\le \text{r}\le \text{n}\].

• By using this formula, any term of the expansion can be calculated.

• MIDDLE TERM \[\left( \text{S} \right)\] :

1. In \[{{\left( \text{a+b} \right)}^{\text{n}}}\]  if \[\text{n}\] is even then the number of terms in the expansion is odd. Therefore there is only one middle term and it is \[{{\left( \dfrac{\text{n}+2}{2} \right)}^{\text{th}}}\] term.

2. In \[{{\left( \text{a+b} \right)}^{\text{n}}}\], if n is odd then the number of terms in the expansion is even. Therefore there are two middle terms and those are \[{{\left( \dfrac{\text{n}+1}{2} \right)}^{\text{th}}}\] and  \[{{\left( \dfrac{\text{n}+3}{2} \right)}^{\text{th}}}\] terms.

Binomial Theorem for any Index:

If \[\text{n}\] is negative integer then \[\text{n!}\] cannot be defined. We state binomial theorem in another form.

\[{{\left( \text{a+b} \right)}^{\text{n}}}\text{=}{{\text{a}}^{\text{n}}}\text{+}\dfrac{\text{n}}{1!}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}\dfrac{\text{n}\left( \text{n}-1 \right)}{2!}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-\text{r}+1 \right)}{\text{r}!}{{\text{a}}^{\text{n-r}}}{{\text{b}}^{\text{r}}}+...\]

Here,

\[{{\text{t}}_{\text{r+1}}}\text{ = }\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-\text{r}+1 \right)}{\text{r}!}{{\text{a}}^{\text{n-r}}}{{\text{b}}^{\text{r}}}\]

Theorem :

• If \[\text{n}\] is any real number,

\[\text{a = 1,b = x}\] and \[\left| \text{x} \right| < 1\] then

\[{{\left( 1+\text{x} \right)}^{\text{n}}}\text{= 1+ nx + }\dfrac{\text{n}\left( \text{n}-1 \right)}{\text{2}!}{{\text{x}}^{2}}+\text{ }\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-2 \right)}{\text{3}!}{{\text{x}}^{3}}+...\]

Here there are infinite number of terms in the expansion, the general term is given by

\[{{\text{t}}_{\text{r+1}}}\text{ = }\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-2 \right)\left( \text{n}-\text{r}+1 \right)}{\text{r}!}\text{, r}\ge \text{0}\]

Note:

1. Expansion is valid only when \[\text{-1 x 1}\]

2. \[^{\text{n}}{{\text{C}}_{\text{r}}}\] can not be used because it is defined only for natural number, so \[^{\text{n}}{{\text{C}}_{\text{r}}}\] can be written as \[\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-2 \right)\left( \text{n}-\text{r}+1 \right)}{\text{r}!}\]

3. As the series never terminates, the number of terms in the series is infinite.

4. General term of the series \[{{\text{(1+x)}}^{-\text{n}}}={{\text{T}}_{\text{r+1}}}\to {{\left( -1 \right)}^{\text{r}}}\]

\[\dfrac{\text{1+x}}{\text{1-x}}\] if \[\left| \text{x} \right| < 1\]

5. General term of the series \[{{\text{(1+x)}}^{-\text{n}}}\to {{\text{T}}_{\text{r+1}}}\]

\[\text{= }\dfrac{\text{   }\left( \text{ -1} \right)\left( \text{ +2} \right)...\left( \text{ +  -1} \right)}{\text{r!}}\text{x}\] 

6. If first term is not $1$, then make it unity in the following way.

\[{{\text{(a+x)}}^{\text{n}}}={{\text{a}}^{\text{n}}}{{\left( \dfrac{\text{1+x}}{\text{a}} \right)}^{\text{n}}}\]if \[\left| \dfrac{\text{x}}{\text{a}} \right| < 1\]

Remarks:

1. If \[\left| \text{x} \right| < 1\] and \[\text{n}\]is any real number, then

\[{{\left( 1-\text{x} \right)}^{\text{n}}}\text{= 1- nx + }\dfrac{\text{n}\left( \text{n}-1 \right)}{\text{2}!}{{\text{x}}^{2}}-\text{ }\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-2 \right)}{\text{3}!}{{\text{x}}^{3}}+...\]

The general term is given by

\[{{\text{t}}_{\text{r+1}}}\text{ = }\dfrac{{{\left( -1 \right)}^{\text{r}}}\text{n}\left( \text{n}-1 \right)\left( \text{n}-2 \right)\left( \text{n}-\text{r}+1 \right)}{\text{r}!}{{\text{x}}^{\text{r}}}\]

2. If \[\text{n}\]is any real number and \[\left| \text{b} \right| < \left| \text{a} \right|\], then

\[{{\text{(a+b)}}^{\text{n}}}={{\left[ \text{a}\left( 1+\dfrac{\text{b}}{\text{a}} \right) \right]}^{\text{n}}}\]

\[{{\text{(a+b)}}^{\text{n}}}={{\text{a}}^{\text{n}}}{{\left( 1+\dfrac{\text{b}}{\text{a}} \right)}^{\text{n}}}\]

Note:

While expanding \[{{\text{(a+b)}}^{\text{n}}}\] where \[\text{n}\]a negative integer or a fraction is, reduce the binomial to the form in which the first term is unity and the second term is numerically less than unity.

Particular expansion of the binomials for negative index, \[\left| \text{x} \right| < 1\]

$\dfrac{\text{1}}{\text{1+x}}\text{ = }{{\left( \text{1+x} \right)}^{\text{-1}}}  \text{        = 1-x+}{{\text{x}}^{\text{2}}}\text{-}{{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{4}}}\text{-}{{\text{x}}^{\text{5}}}\text{+}....$

$\dfrac{\text{1}}{\text{1-x}}\text{ = }{{\left( \text{1-x} \right)}^{\text{-1}}} \text{        = 1+x+}{{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{4}}}\text{+}{{\text{x}}^{\text{5}}}\text{+}....$

$\dfrac{\text{1}}{{{\left( \text{1+x} \right)}^{2}}}\text{ = }{{\left( \text{1+x} \right)}^{\text{-2}}} \text{        = 1-2x+3}{{\text{x}}^{\text{2}}}\text{-4}{{\text{x}}^{\text{3}}}\text{+}.... $

$ \dfrac{\text{1}}{{{\left( \text{1-x} \right)}^{2}}}\text{ = }{{\left( \text{1-x} \right)}^{\text{-2}}} \text{        = 1+2x+3}{{\text{x}}^{\text{2}}}\text{+4}{{\text{x}}^{\text{3}}}\text{+}.... $

Binomial Coefficients:

The coefficients \[{}^{\text{n}}{{\text{C}}_{\text{0}}}\text{,}{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{,}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{, }..\text{ ,}{}^{\text{n}}{{\text{C}}_{\text{n}}}\] in the expansion of \[{{\text{(a+b)}}^{\text{n}}}\] are called the binomial coefficients and denoted as \[{{\text{C}}_{0}},{{\text{C}}_{1}},{{\text{C}}_{2}},..,{{\text{C}}_{\text{n}}}\] respectively.

Now,

\[{{\left( \text{1+x} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{x}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{x}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{x}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n}}}.....\left( \text{i} \right)\]

Put \[\text{x}=1\]

\[{{\left( \text{1+1} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}\]

\[\therefore \text{   }{{\text{2}}^{\text{n}}}\text{ = }{}^{\text{n}}{{\text{C}}_{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}\]

\[\therefore {}^{\text{n}}{{\text{C}}_{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}={{\text{2}}^{\text{n}}}\]

\[\therefore {{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{{\text{C}}_{\text{n}}}={{\text{2}}^{\text{n}}}\]

Therefore, The sum of all binomial coefficients is \[{{\text{2}}^{\text{n}}}\]

Put \[\text{x}=-1\]in equation \[\left( \text{i} \right)\], we get

\[{{\left( \text{1-1} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}-{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{- }.....\text{+}{{\left( -1 \right)}^{\text{n}}}{}^{\text{n}}{{\text{C}}_{\text{n}}}\]

\[\therefore \text{   0 = }{}^{\text{n}}{{\text{C}}_{\text{0}}}-{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{- }.....\text{+}{{\left( -1 \right)}^{\text{n}}}{}^{\text{n}}{{\text{C}}_{\text{n}}}\]

\[\therefore {}^{\text{n}}{{\text{C}}_{\text{0}}}-{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{- }.....\text{+}{{\left( -1 \right)}^{\text{n}}}{}^{\text{n}}{{\text{C}}_{\text{n}}}=\text{0}\]

\[\therefore {}^{\text{n}}{{\text{C}}_{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{+}{}^{\text{n}}{{\text{C}}_{4}}\text{+ }.....={}^{\text{n}}{{\text{C}}_{\text{1}}}+{}^{\text{n}}{{\text{C}}_{3}}+{}^{\text{n}}{{\text{C}}_{5}}+.....\]

\[\therefore {{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}{{\text{C}}_{4}}\text{+ }.....={{\text{C}}_{\text{1}}}+{{\text{C}}_{3}}+{{\text{C}}_{5}}+.....\]

\[{{\text{C}}_{\text{0}}}\text{,}{{\text{C}}_{\text{2}}}\text{,}{{\text{C}}_{4}},...\] are known as even coefficients

\[{{\text{C}}_{1}}\text{,}{{\text{C}}_{3}}\text{,}{{\text{C}}_{5}},...\] are known as odd coefficients

Let, 

\[{{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}{{\text{C}}_{4}}\text{+ }.....={{\text{C}}_{\text{1}}}+{{\text{C}}_{3}}+{{\text{C}}_{5}}+.....=\text{k}\]

Now,

\[{{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{{\text{C}}_{\text{n}}}={{\text{2}}^{\text{n}}}\]

Therefore,

\[\left( {{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}{{\text{C}}_{\text{4}}}\text{+ }..... \right)\text{ = }\left( {{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{3}}}\text{+}{{\text{C}}_{\text{5}}}\text{+}..... \right)\text{ = }{{\text{2}}^{\text{n}}}\]

\[\text{k+k = }{{\text{2}}^{\text{n}}}\]

\[\text{2k = }{{\text{2}}^{\text{n}}}\]

\[\text{k = }\dfrac{{{\text{2}}^{\text{n}}}}{\text{2}}\]

\[\text{k = }{{\text{2}}^{\text{n-1}}}\]

\[\left( {{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}{{\text{C}}_{\text{4}}}\text{+ }..... \right)\text{ = }\left( {{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{3}}}\text{+}{{\text{C}}_{\text{5}}}\text{+}..... \right)\text{ = }{{\text{2}}^{\text{n-1}}}\]

Therefore,

\[\text{The sum of even coefficients = The sum of odd coefficients = }{{\text{2}}^{\text{n-1}}}\]

Properties of Binomial Coefficient:

The coefficients have been omitted for the sake of simplicity.

\[{}^{\text{n}}{{\text{C}}_{\text{0}}}\text{,}{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{,}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{, }..\text{ ,}{}^{\text{n}}{{\text{C}}_{\text{r}}},....,{}^{\text{n}}{{\text{C}}_{\text{n}}}\] are denoted by \[{{\text{C}}_{0}},{{\text{C}}_{1}},{{\text{C}}_{2}},..,{{\text{C}}_{\text{r}}}\text{,}....{{\text{C}}_{\text{n}}}\]

• \[{{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{{\text{C}}_{\text{n}}}={{\text{2}}^{\text{n}}}\]

• \[{{\text{C}}_{\text{0}}}\text{- }{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{- }.....\text{+ }{{\left( -1 \right)}^{\text{n}}}{{\text{C}}_{\text{n}}}=\text{0}\]

• \[{{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}{{\text{C}}_{4}}\text{+ }.....={{\text{C}}_{\text{1}}}+{{\text{C}}_{3}}+{{\text{C}}_{5}}+.....={{\text{2}}^{\text{n-1}}}\]

• \[{}^{\text{n}}{{\text{C}}_{{{\text{r}}_{1}}}}={}^{\text{n}}{{\text{C}}_{{{\text{r}}_{2}}}}\Rightarrow {{\text{r}}_{\text{1}}}\text{=}{{\text{r}}_{\text{2}}}\text{ or }{{\text{r}}_{\text{1}}}\text{+}{{\text{r}}_{\text{2}}}\text{=n}\]

• \[{}^{\text{n}}{{\text{C}}_{\text{r}}}+{}^{\text{n}}{{\text{C}}_{\text{r-1}}}={}^{\text{n+1}}{{\text{C}}_{\text{r}}}\]

• \[\text{r}{}^{\text{n}}{{\text{C}}_{\text{r}}}=\text{n}{}^{\text{n-1}}{{\text{C}}_{\text{r}}}\]

Some Important Results:

1. \[{{\left( \text{1+x} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{x}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{x}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{x}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n}}}\]

Putting \[\text{x}=1\] and \[-1\], we get

\[{{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{{\text{C}}_{\text{n}}}={{\text{2}}^{\text{n}}}\]

\[{{\text{C}}_{\text{0}}}\text{- }{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{- }.....\text{+ }{{\left( -1 \right)}^{\text{n}}}{{\text{C}}_{\text{n}}}=\text{0}\]

2. Differentiating \[{{\left( \text{1+x} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{x}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{x}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{x}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n}}}\] , on both sides, \[\text{n}{{\left( \text{1+x} \right)}^{\text{n -1}}}\]

\[\text{= }{{\text{C}}_{1}}\text{+2}{{\text{C}}_{\text{1}}}\text{x+3}{{\text{C}}_{3}}{{\text{x}}^{2}}\text{+}.....\text{+n}{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n -1}}}.....\left( 1 \right)\]

\[\text{x=1}\]

\[\Rightarrow \text{n}{{\text{2}}^{\text{n-1}}}\text{ = }{{\text{C}}_{\text{1}}}\text{+2}{{\text{C}}_{\text{1}}}\text{+3}{{\text{C}}_{\text{3}}}\text{+}.....\text{+n}{{\text{C}}_{\text{n}}}\]

\[\text{x = -1}\]

\[\Rightarrow \text{0 = }{{\text{C}}_{\text{1}}}\text{- 2}{{\text{C}}_{\text{1}}}\text{+}.....\text{+}{{\left( \text{-1} \right)}^{\text{n -1}}}\text{n}{{\text{C}}_{\text{n}}}\]

Differentiating \[\left( 1 \right)\] again and again we will have different results.

3. Integrating \[{{\left( \text{1 + x} \right)}^{\text{n}}}\], we get

\[\dfrac{{{\left( \text{1+x} \right)}^{\text{n+1}}}}{\text{n+1}}\text{+C=}{{\text{C}}_{\text{0}}}\text{x+}\dfrac{{{\text{C}}_{\text{1}}}{{\text{x}}^{\text{2}}}}{\text{2}}\text{+}\dfrac{{{\text{C}}_{\text{2}}}{{\text{x}}^{\text{3}}}}{\text{3}}\text{+}....\text{+}\dfrac{{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n+1}}}}{\text{n+1}}\]

Where, \[\text{C}\] is a constant.

Put \[\text{x = 0}\], we get

\[\text{C = - }\dfrac{\text{1}}{\left( \text{n+1} \right)}\]

Therefore,

\[\dfrac{{{\left( \text{1+x} \right)}^{\text{n+1}}}\text{-1}}{\text{n+1}}\text{=}{{\text{C}}_{\text{0}}}\text{x+}\dfrac{{{\text{C}}_{\text{1}}}{{\text{x}}^{\text{2}}}}{\text{2}}\text{+}\dfrac{{{\text{C}}_{\text{2}}}{{\text{x}}^{\text{3}}}}{\text{3}}\text{+}....\text{+}\dfrac{{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n+1}}}}{\text{n+1}}.....\left( \text{2} \right)\]

Put \[\text{x = 1}\] in eqn. \[\left( 2 \right)\], we get

\[\dfrac{{{\text{2}}^{\text{n+1}}}\text{-1}}{\text{n+1}}\text{ = }{{\text{C}}_{\text{0}}}\text{+}\dfrac{{{\text{C}}_{\text{1}}}}{\text{2}}\text{+}\dfrac{{{\text{C}}_{\text{2}}}}{\text{3}}\text{+}....\text{+}\dfrac{{{\text{C}}_{\text{n}}}}{\text{n+1}}\]

Put \[\text{x = -1}\] in eqn.\[\left( 2 \right)\], we get

\[\dfrac{\text{1}}{\text{n+1}}\text{ = }{{\text{C}}_{\text{0}}}\text{- }\dfrac{{{\text{C}}_{\text{1}}}}{\text{2}}\text{+}\dfrac{{{\text{C}}_{\text{2}}}}{\text{3}}\text{ - }....\]

Example:

Find the coefficient of \[{{\text{x}}^{4}}\] in the expansion of \[\dfrac{\text{1+ x}}{1-\text{x}}\] if \[\left| \text{x} \right| < 1\]

Ans:

\[\dfrac{\text{1+ x}}{1-\text{x}}=\left( \text{1+ x} \right){{\left( \text{1- x} \right)}^{-1}}\]

\[=\left( \text{1+ x} \right)\left[ 1+\dfrac{\left( -1 \right)}{1!}\left( -\text{x} \right)+\dfrac{\left( -1 \right)\left( -1-1 \right)}{2!}{{\left( -\text{x} \right)}^{2}} +\dfrac{\left( -1 \right)\left( -1-1 \right)\left( -1-2 \right)}{3!}{{\left( -\text{x} \right)}^{3}}+...\infty \right]\]

\[=\left( \text{1+ x} \right)\left( 1+\text{x}+{{\text{x}}^{2}}+{{\text{x}}^{3}}+{{\text{x}}^{4}}+...\infty  \right)\]

\[=\left( 1+\text{x}+{{\text{x}}^{2}}+{{\text{x}}^{3}}+{{\text{x}}^{4}}+...\infty  \right)+\left( \text{x}+{{\text{x}}^{2}}+{{\text{x}}^{3}}+{{\text{x}}^{4}}+...\infty  \right)\]

\[=1+2\text{x}+2{{\text{x}}^{2}}+2{{\text{x}}^{3}}+2{{\text{x}}^{4}}+...\infty \]

\[{{\text{x}}^{4}}=2\]

Hence, the coefficient of \[{{\text{x}}^{4}}\] is \[2\]

Multinomial Expansion:

In the expansion of \[{{\left( {{\text{x}}_{1}}+{{\text{x}}_{2}}+...+{{\text{x}}_{\text{n}}} \right)}^{\text{m}}}\] where \[\text{m,n }\in \text{ N}\] and \[{{\text{x}}_{1}}+{{\text{x}}_{2}}+...+{{\text{x}}_{\text{n}}}\] are independent variables.

We have

1. Total number of terms \[\text{=}{{\text{ }}^{\text{m+n -1}}}{{\text{C}}_{\text{n -1}}}\]

2. Coefficient of \[{{\text{x}}_{\text{1}}}^{^{{{\text{r}}_{\text{1}}}}}\text{ }{{\text{x}}_{\text{2}}}^{^{{{\text{r}}_{\text{2}}}}}\text{ }{{\text{x}}_{\text{3}}}^{{{\text{r}}_{\text{3}}}}....{{\text{x}}_{\text{n}}}^{{{\text{r}}_{\text{n}}}}\]

(Where, \[{{\text{r}}_{\text{1}}}+{{\text{r}}_{2}}+....+{{\text{r}}_{n}}=\text{m}\], \[{{\text{r}}_{\text{i}}}\in \text{N }\cup \left\{ 0 \right\}\])  is \[\dfrac{\text{m!}}{{{\text{r}}_{\text{1}}}\text{! }{{\text{r}}_{\text{2}}}\text{! }...\text{ }{{\text{r}}_{\text{n}}}\text{!}}\]

3. Sum of all the coefficients is obtained by putting all the variables \[{{\text{x}}_{\text{1}}}\] equal to \[1\]

Example:

Find the total number of terms in the expansion of ${{\left( \text{1+a+b} \right)}^{\text{10}}}$ and coefficient of ${{\text{a}}^{\text{2}}}{{\text{b}}^{\text{3}}}$.

Ans:

Total number of terms \[\text{=}{{\text{ }}^{\text{m+n -1}}}{{\text{C}}_{\text{n -1}}}\]

Therefore,

Total number of terms \[\text{=}{{\text{ }}^{\text{10+3 -1}}}{{\text{C}}_{\text{3 -1}}}\]

\[\text{=}{{\text{ }}^{\text{12}}}{{\text{C}}_{\text{2}}}\]

\[=66\]

Coefficient of ${{\text{a}}^{\text{2}}}{{\text{b}}^{\text{3}}}$can be calculated as,

Coefficient of ${{\text{a}}^{\text{2}}}{{\text{b}}^{\text{3}}}$\[=\dfrac{\text{10!}}{\text{2!}\times \text{3!}\times \text{5!}}\]

\[=2520\]

Class 11 Maths Revision Notes for Chapter-8 Binomial Theorem - Free PDF Download

Binomial Expression

The algebraic expression is called binomial, which includes only two terms. It is a polynomial with two terms. It is also known as a sum or difference between two or more monomials. It is the simplest polynomial form.

Binomial Theorem

Here are the necessary formula that you should always try and remember:

(x + y)ⁿ = ⁿCₒxⁿ⁻⁰ y⁰ + ⁿC₁xⁿ⁻¹y¹ + ⁿC₂xⁿ⁻²y² + ………

+ ⁿCᵣx \[^{n-r}\] y\[^{r}\] + ……. + ⁿC\[_{n-1}\] xy\[^{n-1}\] + ⁿC\[_{n}\]x⁰yⁿ

i.e.. (x + y)ⁿ = \[\sum_{r=0}^{n}\] ⁿCᵣ x\[^{n-1}\]y\[^{r}\]   …..(i)

Here ⁿCₒ, ⁿC₁, ⁿC₂,...... ⁿC\[_{n}\] are called binomial coefficients and 

ⁿCᵣ = \[\frac{n!}{r!(n-r)!}\] for 0 ≤ r ≤ n.

This is called the binomial theorem. 

The various terms that you find in the nCx format are known are binomial coefficients.

Properties of the Binomial Expansion

1. Total number of terms in the expansion of (x + a)n is (n + 1). 

2. The sum of the indices of x and a in each term is n.  

3. It is a correct expansion when the terms are complex numbers.

4. Terms that are equidistant from both ends will have coefficients that are equal. These are termed differently - binomial co-efficients.

5. General term in the expansion of (x + c)n is given by Tr + 1 = nCrx n – r ar . 

6. It is important to note that the values first increase and then decrease as you go ahead in the expansion.

7. The coefficient of xr in the expansion of (1+ x)n is nCr.

Middle Term in the Expansion of(1 + x)n

1.  If n is even, then in the expansion of (x + a)n, the middle term is (n/2 + 1)th terms. 

2. If n is odd, then in the expansion of (x + a)n , the middle terms are (n + 1) / 2 th term and (n + 3) / 2 th term.

Greatest Coefficient 

• If n is even, then in (x + a)n , the greatest coefficient is nCn / 2 

• If n is odd, then in (x + a)n , the greatest coefficient is nCn – 1 / 2 or nCn + 1 / 2 both being equal. 

Fun Facts about the Greatest Term

Do you know that there are different ways to find the greatest term? You can do this by finding whether certain parts of the resulting expansion are integers or not, one can find the greatest term for each expansion.  It is important to understand the definition of integers and remembering the greatest term’s formula for each of the cases. 

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