How did Sir Isaac Newton develop and formulate the famous binomial theorem? After completing combination, I have started to read Binomial Theorem. My book only mentioned about Pascal's Triangle. And the formula was then given straightforward. But how did Sir Issac Newton actually develop this theorem?? How did he formulate all these??? Please help. 
 A: About the Binomial Theorem.
When expanding the powers of a binomial by hand and grouping the terms by identical powers, it is not very hard to observe the pattern:
$$(x+y)^0=1$$
$$(x+y)^1=x+y$$
$$(x+y)^2=x^2+2xy+y^2$$
$$(x+y)^3=x^3+3x^2y+3xy^2+y^3$$
$$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$$
$$\dots$$


*

*the expansion is the sum of all $n+1$ possible products of the form $x^iy^j$, with $i+j=n$.

*the shape of Pascal's triangle clearly appears as the number of terms goes linearly increasing.

*the basic recurrence relation between Pascal's numbers appears with the following expansion:


$$\begin{align}(x+y)(x+y)^3
&=x^4+3x^3y+3x^2y^2+xy^3\ +\\
&\ \ \ \ \ \underline{\ \ \ \ +\ \ x^3y+3x^3y^2+3xy^3+y^4}\\
&=x^4+4x^3y+6x^2y^2+4xy^3+y^4
\end{align}$$
You get a new line of coefficients by adding the last line to itself shifted by one position, so that $C_m^{n+1}=C_{m-1}^n+C_m^n$.
All of this is rather straightforward to establish. More interesting is the link with combinatorial analysis.
If you expand $(x+y)^n$ without regrouping the terms,
$$(x+y)^0=1$$
$$\begin{align}(x+y)^1&=x\\
&+y\end{align}$$
$$\begin{align}(x+y)^2&=xx\\
&+xy+yx\\
&+yy\end{align}$$
$$\begin{align}(x+y)^3&=xxx\\
&+xxy+xyx+yxx\\
&+xyy+yxy+yyx\\
&+yyy\end{align}$$
$$\begin{align}(x+y)^4&=xxxx\\
&+xxxy+xxyx+xyxx+yxxx\\
&+xxyy+xyxy+xyyx+yxxy+yxyx+yyxx\\
&+yyxy+yyyx+xyyy+yxyy\\
&+yyyy\end{align}$$
$$\dots$$
you see that there are $2^n$ terms, and you observe that all combinations of $i$ letters $x$ and $j$ letters $y$ (with $i+j=n$) can be grouped as $x^iy^j$, hence
$$C_m^n=\binom nm.$$
At the same time, you see a bell-shaped histogram appear, that prefigures the Gaussian curve.
A: Here, $(x+y)^n$ can be written as $(x+y)(x+y)(x+y)......n..times$
We have to find the coefficient of $x^by^c$ or $a$ in the term $ax^by^c$ in the expansion of $(x+y)^n$ where $(b+c)=n$. So first we will have to find number of ways we can get $x^b$ and we can get it by finding the number of ways we can multiply $(x+y)$, $b$ times which is same as selecting $b$ $(x+y)$ out of $n$ $(x+y)$s which is equal to $n \choose b$. Now we have to find number of ways we can get $y^c$ from the remaining $(n-b)$, $(x+y)$, which is $n-c \choose c$=$c \choose c$=1. 
So the total number of ways we can get $x^by^c$ is $n \choose b$ and since $(b+c)=n$, it is also equal to $n \choose c$.
Pretty simple Proof. 
A: There's actually nothing to prove in the binomial theorem other than that the series developed is well-defined. (I take it we're talking about the cases when the index is not a positive integer, so that we have an infinite series -- and this is the case usually attributed to Newton since the positive integral case had been known since ancient times).
Now Newton did not prove this theorem, but used it, and felt justified by a combination of physical insight (he was trying to work out some plane areas) and blind faith to work out when the series makes sense.
In sum, apart from issues of convergence, the binomial theorem is actually a definition -- namely an extension of the case when the index is a positive integer to more general indices. (As you may know, the former case can be proved by induction.)
As for a proof of the limits within which this generalisation is valid, I think the Norwegian mathematician Niels Abel was the first to supply one. Of course, you may justify that if $|x|\lt 1,$ then $$\sum_{k=0}^{\infty}{\binom{r}{k}x^k}$$ converges, by the ratio test, where $r$ is any real number.
