How did Sir Isaac Newton develop and formulate the famous binomial theorem?

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.