The general form of binomial expression is $a+b$ and the expansion of $(a+b)^{n}$, $n\epsilon N$, is called the binomial theorem for positive integral index.
The binomial theorem enables us to expand any power of a binomial expression.
Development of Binomial Theorem
We know that,
$(a+b)^{0}=1$
$(a+b)^{1}=a+b$
$(a+b)^{2}=a^{2}+2ab+b^{2}$
$(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$
$(a+b)^{4}=a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}$$+b^{4}$
From these expansions, we observe that,
(i) the total number of terms in each expansion is one more than index.
Ex: in the expansion of $(a+b)^{3}$, the number of terms is 4 whereas the index is 3.
(ii) the powers (indices) of first quantity $'a'$ goes on decreasing by 1 whereas, the powers of the second quantity $'b'$ goes on increasing by 1, in successive terms.
(iii) in each term of expansion, the sum of indices of $a$ and $b$ is the same and is equal to the index of $(a+b)$.
Let us write the coefficients of the terms in a table
We can observe that the coefficients form a certain pattern.
(i) each row starts with 1 and ends with 1.
(ii) each coefficient (except the first and last) in a row is the sum of two coefficients in the preceding row.

This pattern (arrangement of numbers) is known as Pascal's Triangle.
Expansions for the higher powers of Binomial can be written using Pascal's triangle.
Ex: expand $(a+b)^{6}$
The row for index 6 is
1 6 15 20 15 6 1
Using this row for coefficients, we get
$(a+b)^{6}=a^{6}+6a^{5}b+15a^{4}b^{2}+20a^{3}b^{3}$$+15a^{2}b^{4}+6ab^{5}+b^{6}$

Now, if we want to find the expansion of $(a+b)^{12}$, we are required to get the row for index 12, by writing Pascal's triangle till index 12. This is a lengthy process.
Therefore, we use the concept of combinations to rewrite the numbers in the Pascal's triangle.
We know that,
$\displaystyle \small ^{n}C_{r}=\frac{n!}{r!(n-r)!}$, $0\leq r\leq n$
Also,
 $\displaystyle \small ^{n}C_{0}=1=^{n}C_{n}$
The Pascal's triangle can be rewritten as
Now we can write the row of Pascal's triangle for any index without writing the earlier rows.
Ex: for index 7, the row would be
$^{7}C_{0}$, $^{7}C_{1}$, $^{7}C_{2}$, $^{7}C_{3}$, $^{7}C_{4}$, $^{7}C_{5}$, $^{7}C_{6}$, $^{7}C_{7}$
Using this row we can expand $(a+b)^{7}$ as
$\displaystyle \small (a+b)^{7}=^{7}C_{0}a^{7}+^{7}C_{1}a^{6}b+^{7}C_{2}a^{5}b^{2}$$\displaystyle \small +^{7}C_{3}a^{4}b^{3}+^{7}C_{4}a^{3}b^{4}+^{7}C_{5}a^{2}b^{5}$ $\displaystyle \small +^{7}C_{6}ab^{6}+^{7}C_{7}b^{7}$
Binomial theorem for any positive integer n
$\displaystyle \small (a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b+^{n}C_{2}a^{n-2}b^{2}$ $\displaystyle \small +...+^{n}C_{n-1}ab^{n-1}+^{n}C_{n}b^{n}$
Proof:
Let $\displaystyle \small P(n)=(a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b$ $\displaystyle \small +^{n}C_{2}a^{n-2}b^{2}+...$$\displaystyle \small +^{n}C_{n-1}ab^{n-1}+^{n}C_{n}b^{n}$
for $n=1$,
$\displaystyle \small P(1): (a+b)^{1}=^{1}C_{0}a^{1}+^{1}C_{1}b^{1}$
$P(1): (a+b)^{1}=a+b$
Thus $P(1)$ is true.
Suppose $P(k)$ is true for some positive integer $k$,
$\displaystyle \small P(k)=(a+b)^{k}=^{k}C_{0}a^{k}+^{k}C_{1}a^{k-1}b$ $\displaystyle \small +^{k}C_{2}a^{k-2}b^{2}+...$$\displaystyle \small +^{k}C_{k-1}ab^{k-1}+^{k}C_{k}b^{k}$
We shall prove that $P(k+1)$ is true,
$\displaystyle \small P(k+1)=(a+b)^{k+1}= ^{k+1}C_{0}a^{k+1}$ $\displaystyle \small +^{k+1}C_{1}a^{k+1-1}b+^{k+1}C_{2}a^{k+1-2}b^{2}+...$ $\displaystyle \small +^{k+1}C_{k+1-1}ab^{k+1-1}+^{k+1}C_{k+1}b^{k+1}$
Now,
$P(k+1)=(a+b)(a+b)^{k}$
$\displaystyle \small P(k+1)=(a+b)(^{k}C_{0}a^{k}+^{k}C_{1}a^{k-1}b$ $\displaystyle \small +^{k}C_{2}a^{k-2}b^{2}+...$ $\displaystyle \small +^{k}C_{k-1}ab^{k-1}+^{k}C_{k}b^{k})$
$\displaystyle \small P(k+1)=(^{k}C_{0}a^{k}.a+^{k}C_{1}a^{k-1}b.a$ $\displaystyle \small +^{k}C_{2}a^{k-2}b^{2}.a+...$ $\displaystyle \small +^{k}C_{k-1}ab^{k-1}.a+^{k}C_{k}b^{k}.a)$ $\displaystyle \small +(^{k}C_{0}a^{k}.b+^{k}C_{1}a^{k-1}b.b+^{k}C_{2}a^{k-2}b^{2}.b+...$$\displaystyle \small +^{k}C_{k-1}ab^{k-1}.b+^{k}C_{k}b^{k}.b)$
$\displaystyle \small P(k+1)=(^{k}C_{0}a^{k+1}+^{k}C_{1}a^{k}b+^{k}C_{2}a^{k-1}b^{2}$ $\displaystyle \small +...$ $\displaystyle \small +^{k}C_{k-1}a^{2}b^{k-1}+^{k}C_{k}ab^{k})$ $\displaystyle \small +(^{k}C_{0}a^{k}b+^{k}C_{1}a^{k-1}b^{2}+^{k}C_{2}a^{k-2}b^{3}+...$$\displaystyle \small +^{k}C_{k-1}ab^{k}+^{k}C_{k}b^{k+1})$
$\displaystyle \small P(k+1)=^{k}C_{0}a^{k+1}+(^{k}C_{1}+^{k}C_{0})a^{k}b$ $\displaystyle \small +(^{k}C_{2}+^{k}C_{1})a^{k-1}b^{2}+...$ $\displaystyle \small +(^{k}C_{k-1}+^{k}C_{k})ab^{k}+^{k}C_{k}b^{k+1})$
[$\displaystyle \small ^{k+1}C_{0}=1, ^{k}C_{r}+^{k}C_{r-1}=^{k+1}C_{r}$ and $\displaystyle \small ^{k}C_{k}=1=^{k+1}C_{k+1}$]
$\displaystyle \small P(k+1)=^{k+1}C_{0}a^{k+1}+^{k+1}C_{1}a^{k}b$ $\displaystyle \small +^{k+1}C_{2}a^{k-1}b^{2}+...$ $\displaystyle \small +^{k+1}C_{k-1}ab^{k}+^{k+1}C_{k+1}b^{k+1})$

Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Therefore, by principle of mathematical induction, $P(n)$ is true for every positive integer $n$.
Observations:
1. The total number of terms in the expansion of $(a+b)^{n}$ is $(n+1)$ i.e. one more than the index $n$.
2. The sum of indices of $a$ and $b$ in each term is $n$. In the first term of expansion of $(a+b)^{n}$, the index of $a$ starts with $n$, goes on decreasing by 1 in every successive term and ends with 0, whereas the index of $b$ starts with 0, goes on increasing by 1 in every successive term and ends with $n$.
3. The coefficients $^{n}C_{0}$, $^{n}C_{1}$,... are called binomial coefficients.
Some special cases in the expansion of $(a+b)^{n}$
(i) $a=x$ and $b=-y$
$(x-y)^{n}=(x+(-y))^{n}$$\displaystyle \small =^{n}C_{0}x^{n}+^{n}C_{1}x^{n-1}(-y)+^{n}C_{2}x^{n-2}(-y)^{2}+...$ $\displaystyle \small +^{n}C_{n-1}x(-y)^{n-1}+^{n}C_{n}(-y)^{n}$
= $\displaystyle \small ^{n}C_{0}x^{n}-^{n}C_{1}x^{n-1}y+^{n}C_{2}x^{n-2}y^{2}+...$ $\displaystyle \small +(-1)^{n}\times ^{n}C_{n}y^{n}$ 

(ii) $a=1$, $b=x$
$\displaystyle \small (1+x)^{n}=^{n}C_{0}1^{n}+^{n}C_{1}1^{n-1}x+^{n}C_{2}1^{n-2}x^{2}+...$ $\displaystyle \small +^{n}C_{n-1}1x^{n-1}+^{n}C_{n}x^{n}$
= $\displaystyle \small ^{n}C_{0}+^{n}C_{1}x+^{n}C_{2}x^{2}+...$ $\displaystyle \small +^{n}C_{n-1}x^{n-1}$ $\displaystyle \small +^{n}C_{n}x^{n}$  

(iii) $a=1$, $b=-x$
$\displaystyle \small (1-x)^{n}=^{n}C_{0}1^{n}+^{n}C_{1}1^{n-1}(-x)$ $\displaystyle \small +^{n}C_{2}1^{n-2}(-x)^{2}$ $\displaystyle \small +...$ $\displaystyle \small +^{n}C_{n-1}1(-x)^{n-1}+^{n}C_{n}(-x)^{n}$
= $\displaystyle \small ^{n}C_{0}-^{n}C_{1}x+^{n}C_{2}x^{2}-...+(-1)^{n}\times ^{n}C_{n}x^{n}$