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Q9RE

Expert-verifiedFound in: Page 440

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**a) What is Pascal’s triangle? **

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

(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 $\left(\left.\begin{array}{c}\mathrm{n}\\ \mathrm{k}-1\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}\end{array}\right.\right)$ .

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

(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 $\mathrm{n}\ge \mathrm{k},\left(\left.\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}-1\end{array}\right.\right)$ .

(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 –

${T}_{k}(n+1)={T}_{k-1}\left(n\right)+{T}_{k}\left(n\right)\phantom{\rule{0ex}{0ex}}=\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{c}n\\ k\end{array}\right)\phantom{\rule{0ex}{0ex}}=\left(\begin{array}{c}n+1\\ k\end{array}\right)$

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

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