Question

a) What is Pascal’s triangle?

b) How can a row of Pascal’s triangle be produced from the one above it?

Short answer

(a) A geometric arrangement of the binomial coefficients in a triangle which is based on Pascal's identity$n\ge k,\left(\left.\begin{array}{c}n+1\\ k\end{array}\right)=\left(\begin{array}{c}n\\ k\end{array}\right)+\left(\begin{array}{c}n\\ k-1\end{array}\right.\right)$ is known as Pascal’s triangle.

(b) A row of Pascal’s triangle can be produced from the one above it by using the formula$\left(\left.\begin{array}{c}\mathrm{n}\\ \mathrm{k}-1\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}\end{array}\right.\right)$ .

Step-by-step solution

Concept Introduction

Pascal's triangle is a triangular array of binomial coefficients found in probability theory, combinatorics, and algebra in mathematics.

Pascal’s Triangle

(a)

Pascal's triangle is a geometric arrangement of the binomial coefficients in a triangle based on the principle called Pascal's identity which states that for$\mathrm{n}\ge \mathrm{k},\left(\left.\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}-1\end{array}\right.\right)$. The row in the triangle consists of the binomial coefficients –

$\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right),\mathrm{k}=0,1,2,\dots ,\mathrm{n}$

Therefore, Pascal triangle is based on Pascal’s identity$\mathrm{n}\ge \mathrm{k},\left(\left.\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}-1\end{array}\right.\right)$ .

Row of Pascal’s Triangle

(b)

The${\mathrm{k}}^{\mathrm{th}}$ term in the${\left(\mathrm{n}+1\right)}^{\mathrm{th}}$ row in Pascal's triangle is just the sum of the${\left(\mathrm{k}-1\right)}^{\mathrm{th}}$ and${\mathrm{k}}^{\mathrm{th}}$ terms in its above row, i.e., the${\mathrm{n}}^{\mathrm{th}}$ row. Let${\mathrm{T}}_{\mathrm{k}}\left(\mathrm{n}\right)$be the${\mathrm{k}}^{\mathrm{th}}$ term in the${\mathrm{n}}^{\mathrm{th}}$ row of Pascal's triangle for$\mathrm{k}=0,1,2,...,\mathrm{n},$ then the${\left(\mathrm{n}+1\right)}^{\mathrm{th}}$ row can be obtained as –

$$\begin{gathered} T_k(n+1)=T_{k-1}\left(n\right)+T_k\left(n\right) \\ =\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{c}n\\ k\end{array}\right) \\ =\left(\begin{array}{c}n+1\\ k\end{array}\right) \end{gathered}$$

Therefore, the row can be obtained using the formularole="math" localid="1668683483371" $\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}-1\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}\end{array}\right)$ .