- An algebraic expression containing two terms is called binomial expression.

Example

$( x+a)$

$(\frac {1}{2} +x)$

$(\frac {2}{x} - \frac{1}{x^3})$ - The general form of the binomial expression is $(x+a)$ and expansion $(x+a)^n , n \in N$ is called Binomial expression
- It was developed by Sir Issac newton
- The general expression for the Binomial Theorem is

$(x+a)^n = ^{n}C_{0} x^n a^0 + ^{n}C_{1} x^{n-1} a^1 + ^{n}C_{2} x^{n-2} a^2 + .....+ ^{n}C_{r} x^{n-r} a^r ....+^{n}C_{n} x^{0} a^n $

$(x+a)^{n} = \sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^{k} $

**Proof:**

We can prove this theorem with the help of mathematical induction

Let us assume P(n) be the statement is

$(x+a)^n = ^{n}C_{0} x^n a^0 + ^{n}C_{1} x^{n-1} a^1 + ^{n}C_{2} x^{n-2} a^2 + .....+ ^{n}C_{r} x^{n-r} a^r ....+ ^{n}C_{n} x^{0} a^n $

**Step 1**

Now the value of P(1)

$(x+a)^1 = ^{1}C_{0} x^1 a^0 + ^{1}C_{1} x^{1-1} a^1$

$=(x+a)$

So P(1) is true

**Step 2**

Now the value of P(m)

$(x+a)^m = ^{m}C_{0} x^m a^0 + ^{m}C_{1} x^{m-1} a^1 + ^{m}C_{2} x^{m-2} a^2 + ...+ ^{m}C_{m} x^{0} a^m $

Now we have to prove

$(x+a)^{m+1} = ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 +...+ ^{m+1}C_{m+1} x^{0} a^{m+1} $

Now

$(x+a)^{m+1} =(x+a)(x+a)^m $

$= (x+a)( ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 + ...+ ^{m+1}C_{m+1} x^{0} a^{m+1} $

$= ^m C_0 x^{m+1}a{0} + ( ^mC_1 + ^mC_0) x^ma^1 + (^mC_2 + ^mC_1) x^{m-1}a^2 +.... $ $+(^m C_{m-1} + ^mC_{m}) x^1a^m + ^mC_m x^0a^{m+1}$

As $ ^mC_{r-1} + ^mC_r = ^{m+1}C_r$

So,

$= ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 + ...+ ^{m+1}C_{m+1} x^{0} a^{m+1} $

So by principle of Mathematical induction, P(n) is true for $n \in N$

1 |
$(x+a)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^{k} $ We can easily see that $(x+a)^n$ has $(n+1)$ terms as k can have values from 0 to n |

2 |
The sum of indices of x and a in each is equal to n $x^{n-k}a^{k}$ |

3 |
The coefficient ^{n}C_{r} is each term is called binomial coefficient |

4 |
$(x-a)^n$ can be treated as $[x+(-a)]^n$ So $(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k $ So the terms in the expansion are alternatively positive and negative. The last term is positive or negative depending on the values of n |

5. |
$(1+x)^n$ can be treated as $(x+a)^1$ where x=1 and a=x So $(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$ This is the expansion is ascending order of power of x |

6. |
$(1+x)^n$ can be treated as $(x+a)^n$ where x=x and a=1 So $(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}$ This is the expansion is descending order of power of x |

7. |
$(1-x)^n$ ;can be treated as $(x+a)^n$ where x=1 and a=-x So $(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$ This is the expansion is ascending order of power of x |

Expand using Binomial Theorem

$(x^2 +2)^5$

We know from binomial Theorem

$(x+a)^n = ^{n}C_{0} x^n a^0 + ^{n}C_{1} x^{n-1} a^1 + ^{n}C_{2} x^{n-2} a^2 + .....+ ^{n}C_{r} x^{n-r} a^r ....+^{n}C_{n} x^{0} a^n $

So putting values $x=x^2 ,a=2 \; and \; n=5 $

We get

$(x^2 +2)^5= ^{5}C_{0} (x^2)^5 + ^{5}C_{1} (x^2)^4 2^1 + ^{5}C_{2} (x^2)^3 2^2 + ^{5}C_{3} (x^2)^2 2^3 + ^{5}C_{4} (x^2)^1 2^4 + ^{5}C_{5} (x^2)^0 2^5$ $=x^10 +20x^8 +160x^6 +640x^4 +1280x^2 +1024$

- $(x +2)^6$
- $(1 -x^2)^5$
- $(1 -x)^7$
- $(z - x)^5$

The First term would be = $ ^{n} C_0 x^n a^0$

The Second term would be =$ ^{n} C_1 x^{n-1} a^1$

The Third term would be = $ ^{n} C_2 x^{n-2} a^2$

The Fourth term would be = $ ^{n} C_3 x^{n-3} a^3$

Like wise (k+1) term would be

$T_{k+1}= ^n C_k x^{n-k}a^k $

This is called the general term also as every term can be find using this term

$T_1= T_{0+1}=^n C_0 x^n a^0 $

$T_2= T_{1+1}=^n C_1 x^{n-1} a^1 $

Similarly for

$(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k $

$T_{k+1}= ^n C_k (-1)^k x^{n-k}a^k $

Again similarly for

$(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$

$T_{k+1}= ^n C_k x^{k}$

Again similarly for

$(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$

$T_{k+1}= ^n C_k (-1)^k x^{k}$

Binomial term |
(k+1) term |

$(x+a)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^k $ |
$T_{k+1}= ^n C_k x^{n-k}a^k $ |

$(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k $ |
$T_{k+1}= ^n C_k (-1)^k x^{n-k}a^k $ |

$(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$ |
$T_{k+1}= ^n C_k x^{k}$ |

$(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$ |
$T_{k+1}= ^n C_k (-1)^k x^{k}$ |

- A binomial expansion contains $(n+1)$ terms
- If n is even then the middle term would $[(\frac {n}{2}+1 ]$ th term
- If n is odd,then $\frac{n+1}{2} $ and $ \frac {n+3}{3}$ are the middle term

If the coefficient of $(2k + 4)$ and $(k - 2)$ terms in the expansion of $(1+x)^{24}$ are equal then find the value of k

The general term of $(1 + x)^n$ is

$T_{k+1}= ^n C_k x^{k}$

Hence coefficient of (2k + 4)

$T_{2k+4} = T_{2k+3+1} = ^{24} C_{2k+3}$

and coefficient or (k - 2)

$T_{k-2} = T_{k-3+1} = ^{24} C_{k-3}$

As per question both the terms are equal

or (2k + 3) + (k-3) = 24

k = 8

- Prove that $^nC_0 + ^nC_1 + ^nC_2 + .....+ ^nC_n = 2^n $
- Find the Coefficent of $x^5$ in the expansion $(1+x)^3 (1-x)^6$
- Use Binomial theorem to evaluate $(96)^3$

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