### Binomial Theorem Class XI Chapter 8

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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**

**F**

**irst 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)^{1 }= a + b

(a + b)^{2}= a^{2}+2ab +b^{2}

(a + b)^{3}= a^{3 }+ 3a^{2}b^{ }+ 3ab^{2 }+ b^{3}

(a + b)^{4}= a^{4 }+ 4a^{3}b + 6a^{2}b^{2 }+ 4ab^{3 }+ b^{4}

^{}

**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.
- All rows start with 1 and ends with 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.a^{2}+ 2.a + 1.a^{0}

Write increasing order of b (start with 0)

1.a^{2}.b^{0 }+ 2 . a . b + 1 . a^{0 }. b^{2}

Therefore we have (a + b)^{2}= a^{2}+ 2ab + b^{2}

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

1a^{3}+ 3a^{2}+ 3a^{1}+ 1a^{0}

Write increasing order of b (start with 0)

1a^{3}b^{0}+ 3a^{2}b^{1}+ 3a^{1}b^{2}+ 1a^{0}b^{3}

Therefore we have (a + b)^{3}= a^{3}+ 3a^{2}b + 3ab^{2}+ b^{3}

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.a^{6}.b^{0}+ 6.a^{5}b^{1}+ 15.a^{4}.b^{2}+ 20.a^{3}.b^{3}+ 15.a^{2}.b^{4}+ 6.a^{1}.b^{5}+ 1.a^{0}.b^{6}

(a + b)^{6}= a^{6}+ 6a^{5}b+15a^{4}b^{2}+ +20.a^{3}.b^{3}+ 15.a^{2}.b^{4}+ 6.a.b^{5}+ b^{6}

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

+ - + - + - + ……..

(a - b)^{6}= a^{6}- 6a^{5}b + 15a^{4}b^{2}- 20.a^{3}.b^{3}+ 15.a^{2}.b^{4}- 6.a.b^{5}+ b^{6}

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 r^{th }term from the end of the expansion

Let index = n

Total terms of the expansion is = n + 1

r^{th}term from the end of the expansion is = Total terms - r + 1

= n + 1 - r + 1 = (n - r + 2)^{th}term from the startingTherefore r^{th}term from the end of the expansion is the (n - r + 2)^{th}term from the starting.

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