An algebraic expression containing two terms is called binomial expression.
Example
( x+a)
(1/2 +x)
(2/x 1/x^{3})
The general form of the binomial expression is (x+a) and expansion (x+a)^{n} n? 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^{n1}a^{1} + ^{n}C_{2} x^{n2}a^{2} +………………+ ^{n}C_{r} x^{nr}a^{r} …….+ ^{n}C_{n} x^{0}a^{n}
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^{n1}a^{1} + ^{n}C_{2} x^{n2}a^{2} +………………+ ^{n}C_{r} x^{nr}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^{11}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^{m1}a^{1} + ^{m}C_{2} x^{m2}a^{2} +………………+ ^{m}C_{r} x^{mr}a^{r} …….+ ^{m}C_{n} 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^{m1}a^{2} +………………+ ^{m+1}C_{r} x^{mr1}a^{r} …….+ ^{m+1}C_{m+1} x^{0}a^{m+1}
Now
(x+a)^{m+1} =(x+a)(x+a)^{m}
= (x+a)(^{m}C_{0} x^{m}a^{0} + ^{m}C_{1} x^{m1}a^{1} + ^{m}C_{2} x^{m2}a^{2} +………………+ ^{m}C_{r} x^{mr}a^{r} …….+ ^{m}C_{n} x^{0}a^{m})
=^{ m}C_{0} x^{m+1}a^{0} + ( ^{m}C_{1} +^{ m}C_{0}) x^{m}a^{1} + ( ^{m}C_{2} +^{ m}C_{1}) x^{m1}a^{2} + ………
+( ^{m}C_{m1} +^{ m}C_{m}) x^{1}a^{m} + ^{m}C_{m} x^{0}a^{m+1}
As ^{m}C_{r1} +^{ m}C_{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^{m1}a^{2} +………………+ ^{m+1}C_{r} x^{mr1}a^{r} …….+ ^{m+1}C_{m+1} x^{0}a^{m+1}
So by principle of Mathematical induction, P(n) is true for n? N
1 
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^{nk}a^{k} 
3 
The coefficient ^{n}C_{r} is each term is called binomial coefficient 
4 
(xa)^{n } can be treated as [x+(a)]^{n} So 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)^{n} where x=1 and a=x So 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 This is the expansion is descending order of power of x 
7. 
(1x)^{n } can be treated as (x+a)^{n} where x=1 and a=x So This is the expansion is ascending order of power of x 
Examples: Expand using Binomial Theorem
(x^{2} +2)^{5}
Solution: We know from binomial Theorem
(x+a)^{n}= ^{n}C_{0} x^{n}a^{0} + ^{n}C_{1} x^{n1}a^{1} + ^{n}C_{2} x^{n2}a^{2} +………………+ ^{n}C_{r} x^{nr}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
The First term would be =^{ n}C_{0} x^{n}a^{0}
The Second term would be =^{ n}C_{1} x^{n1}a^{1}
The Third term would be =^{ n}C_{2} x^{n2}a^{2}
The Fourth term would be =^{ n}C_{3} x^{n3}a^{3}
Like wise (k+1) term would be
T_{k+1}= ^{n}C_{k} x^{nk}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^{n1}a^{1}
Similarly for
T_{k+1}= ^{n}C_{k} (1)^{k} x^{nk}a^{k}
Again similarly for
T_{k+1}= ^{n}C_{k} x^{k}
Again similarly for
T_{k+1}= ^{n}C_{k} (1)^{k} x^{k}
To summarize it
Binomial term 
(k+1) term 

T_{k+1}= ^{n}C_{k} x^{nk}a^{k}


T_{k+1}= ^{n}C_{k} (1)^{k} x^{nk}a^{k}


T_{k+1}= ^{n}C_{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 [(n/2)+1 ]th term
If n is odd,then (n+1)/2 and (n+3)/3 are the middle term
Examples:
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
Solution:
The general term of (1 + x)^{n} is
T_{k+1}= ^{n}C_{k} x^{k}
Hence coefficient of (2k + 4)^{th} term will be
T_{2k+4} = T_{2k+3+1} = ^{24}C_{2k+3}
and coefficient or (k  2)^{th} term will be
T_{k2} = T_{k3+1} = ^{24}C_{k3}
As per question both the terms are equal
^{24}C_{2k+3} = ^{24}C_{k3}.
=> (2k + 3) + (k3) = 24
.·. r = 8