### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

### Binomial Theorem Class XI Chapter 8

Binomial Theorem Class 11 Chapter 8
Basic concepts related to Binomial theorem chapter 8 class 11, Pascal's method of finding the coefficients, Expansion of Binomial Theorem, Its general term and total number of terms

Binomial:
Algebraic expression which contains only two terms is called binomial.
In this blog we will discuss the followings:
• Method of expanding the binomials with different powers(Indexes).
• Explanation through Pascal's Triangle.
• Finding the coefficients by using Combinations.
• Total terms in the expansion of binomials.
• Expansion of Binomial Theorem.
• General term of Binomial Theorem.
Method of finding the Algebraic Identities
First of all we want to explain the method of expanding the binomials with different powers. Powers of the binomial is called index of the binomial.
Let us  see the following identities.
(a + b)0   = 1

(a + b)=  a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3   = a+ 3a2b  + 3ab + b3

(a + b)4 = a+ 4a3b + 6a2b+ 4ab+ b4

What we observe from these identities?
• Power of 'a' start with the index and goes on decreasing up to zero.
• Power of 'b' start with 0 and goes up to the index.
• At all the terms the sum of powers of 'a' and 'b' always equal to the index 'n'.
• Number of terms is one more than the index.
• Each term of the expansion have some coefficients.
Method of finding the coefficients
Coefficients of any binomial can be calculate with the help of Pascal's Triangle

First of all try to understand how we make Pascal's Triangle
• In the First row the term is  1, It is for the index 0. Because when index is 0, then number of terms  is 1.
• In the II row the  terms are  1, 1. It is for the index 1. Because when index is 1 then number of terms two.
• In the III row the  terms are  1,2, 1. It is for the index 2. Because when index is 2 then number of terms three.
• Number of terms in any expansion is always one more than the index.
• For index n, the number of terms is n + 1.
Explanation

The symbol shown here means that  sum of the left and

right term of triangle equal to the bottom term.

i. e.     1 + 2 = 3
The Numbers written in this given figure means that .
5 + 20 = 25
Power of the binomial is called index.

For example :

Index of   (a + b)2 is 2,   Index of   (a + b)3 is 3 and so on.

For index 2 we have 3 coefficients: 1, 2, 1

For index 3 we have 4 coefficients : 1, 3, 3, 1

For index 4 we have 5 coefficients: 1, 4, 6, 4, 1 and so on.

Now Expand   (a + b)2

Here index is = 2

Write the coefficients from the Pascal’s Triangle

1      +       2      +     1

Write decreasing order of powers of a

1.a2  +   2.a     +    1.a0

1.a2 .b0  +  2 . a . b  +  1 . a. b2

Therefore we have    (a + b)2   = a2  +  2ab  + b2

Expand  (a + b)3

Here index is = 3

Write the coefficients from the Pascal’s Triangle

1      +       3      +     3     +    1

Write decreasing order of powers of a

1a3   +       3a2  +    3a1  +   1a0

1a3 b0  +    3a2 b1 +    3a1 b2 +   1a0 b3

Therefore we have  (a + b)3  = a3 + 3a2b + 3ab2 + b3

Algorithms for making the identities

·         Power of 'a' start with the index and goes on decreasing up to zero

·         Power of 'b' start with 0 and goes up to the index.

·         At all the terms the sum of powers of a and b always equal to the index.

·         Coefficients for all the terms can be written with the help of Pascal’s triangle.

By using above algorithm we can expand the identity (a + b)6 as follows

Index = 6

So total number of terms should be  6 + 1 = 7

Coefficients  from Pascal's triangle are for index 6:

1  +   6  +  15  +  20  +  15  +  6   +   1

Write powers of a and b according to the method mentioned above we get

(a +b)6  =  1.a6.b0  +  6.a5b1 + 15.a4.b2 + 20.a3.b3 + 15.a2.b4 + 6.a1.b5 + 1.a0.b6

(a + b)6 = a6 + 6a5b+15a4b2+ +20.a3.b3 + 15.a2.b4 + 6.a.b5 + b6

Similarly we can expand (a -b)6 by simply using sign as

+   -   +   -   +   -   +  ……..

(a - b)6 = a6 - 6a5b + 15a4b2 - 20.a3.b3 + 15.a2.b4 -  6.a.b5 + b6

Coefficients of the binomial expansion can also be calculated with the help of combinations.$^{6}C_{0}=1,\: \: ^{6}C_{1}=6,\: \: ^{6}C_{2}=15,\\ ^{6}C_{3}=20,\: \: ^{6}C_{4}=15,\: \: ^{6}C_{5}=6,\: \: ^{6}C_{6}=1$Combinations can be calculated by using the formula$^{n}C_{r}=\frac{n!}{r!(n-r)!}$When we expand any binomial by using the combinations for finding the coefficients then we get Binomial Theorem.

Binomial Theorem:

It is a theorem used to expand any binomial with any index by using the combinations.

Expansion of binomial theorem is given below.

$\left ( a+b \right )^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b+^{n}C_{2}a^{n-2}b^{2}+^{n}C_{3}a^{n-3}b^{3}+\\.....+ ^{n}C_{r}a^{n-r}b^{r}+........+ ^{n}C_{n}a^{0}b^{n}$

General term of the binomial is given by $T_{r+1}= ^{n}C_{r}a^{n-r}b^{r},\: where\: 0\leq r\leq n$Total terms in the binomial is = n+1

Power of a starts from n and continuously goes on decreasing to 0

Power of b starts from 0 and continuously goes on increasing to n

OR

$\left ( a-b \right )^{n}=^{n}C_{0}a^{n}b^{0}-^{n}C_{1}a^{n-1}b^{1}+^{n}C_{2}a^{n-2}b^{2}-^{n}C_{3}a^{n-3}b^{3}+\\......+ \left ( -1 \right )^{r}\times \: ^{n}C_{r}a^{n-r}b^{r}.........\left ( -1 \right )^{n}\times ^{n}C_{n}a^{0}b^{n}$General term of this expansion is $T_{r+1}=\: \:\left ( -1 \right )^{r}\times \: ^{n}C_{r}a^{n-r}b^{r},\: where\: 0\leq r\leq n$

To find the rth  term from the end of the expansion

Let index = n

Total terms of the expansion is = n + 1

rth term from the end of the expansion is  = Total terms - r + 1

=  n + 1 - r + 1 = (n - r + 2)th  term from the starting

Therefore rth  term from the end of the expansion is the (n - r + 2)th term from the starting.