# Newton's binomial and Pascal's triangle

## Newton's binomial

Newton's binomial is an algorithm that allows to calculate any power of a binomial; to do so we use the binomial coefficients, which are only a succession of combinatorial numbers. The general formula of Newton's binomial states:

$$(a+b)^n = \begin{pmatrix} n \\ 0 \end{pmatrix} a^n + \begin{pmatrix} n \\ 1 \end{pmatrix} a^{n-1} b + \begin{pmatrix} n \\ 2 \end{pmatrix} a^{n-2} b^2 + \ldots +$$$$$\begin{pmatrix} n \\ n-1 \end{pmatrix} a b^{n-1} + \begin{pmatrix} n \\ n \end{pmatrix} b^{n}$$$

The combinatorial numbers that appear in the formula are called binomial coefficients.

For instance:

$$\begin{array}{rl} (a+b)^4 =& \begin{pmatrix} 4 \\ 0 \end{pmatrix} a^4 + \begin{pmatrix} 4 \\ 1 \end{pmatrix} a^3 b + \begin{pmatrix} 4 \\ 2 \end{pmatrix} a^2 b^2 + \begin{pmatrix} 4 \\ 3 \end{pmatrix} a b^3 + \begin{pmatrix} 4 \\ 4 \end{pmatrix} b^4 \\ =& a^4+4a^3b+6a^2b^2+4ab^3+b^4 \end{array}$$$(In the case where in the binomial there is a negative sign, the signs of the development have to alternate as follows $$+ \ -\ +\ -\ +\ -\ \ldots$$) ## Pascal's triangle Pascal designed a simple way to calculate combinatorial numbers (although this idea is attributed to Tartaglia in some texts): $$\begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1 & & 1 & & & & \\ & & & 1 & & 2 & & 1 & & & \\ & & 1 & & 3 & & 3 & & 1 & & \\ & 1 & & 4 & & 6 & & 4 & & 1 & \\ 1& & 5 & & 10 & & 10 & & 5 & & 1 \end{array}$$$

The method receives the name of triangle of Pascal and is constructed of the following form (fin lines and from top to bottom):

• In the apex a $$1$$ is placed.
• Every line begins and ends in $$1$$.
• The other numbers of the line are always the sum of the two numbers above.

The last line, for example, would give us the value of the consecutive combinatorial numbers:

$$\begin{pmatrix} 5 \\ 0 \end{pmatrix}, \quad \begin{pmatrix} 5 \\ 1 \end{pmatrix}, \quad \begin{pmatrix} 5 \\ 2 \end{pmatrix}, \quad \begin{pmatrix} 5 \\ 3 \end{pmatrix}, \quad \begin{pmatrix} 5 \\ 4 \end{pmatrix}, \quad \begin{pmatrix} 5 \\ 5 \end{pmatrix}$$$The general term of the development of $$(a+b)^n$$ is given by the formula: $$\begin{pmatrix} n \\ k \end{pmatrix} a^{n-k}b^k$$$

According to this, in the previous example we would have the third term would be (for $$k = 2$$, since the series always begins with $$k = 0$$):

$$\begin{pmatrix} 4 \\ 2 \end{pmatrix} a^2 b^2=6a^2b^2$$$This formula allows us to calculate the value of any term without carrying the whole development out. To calculate the 20th term of the development of $$(x+y)^{30}$$. Applying the formula: $$\begin{pmatrix} 30 \\ 19 \end{pmatrix} x^{30-19} y^{19} = 54627300 x^{11}y^{19}$$$