Question

How do one use Pascal’s triangle to expand the binomial ${{\left( d-5 \right)}^{6}}$?

Answer 1

Hint: We first write the binomial expansion of ${{\left( d-5 \right)}^{6}}$. Then we explain Pascal's triangle and the use of it. We explain how the coefficients work. We also explain the use of the constant a and n in the general expansion of ${{\left( x+a \right)}^{n}}$. Then we find the coefficients of the equation ${{\left( d-5 \right)}^{6}}$ using Pascal’s triangle.

Complete step by step answer:
First, we write down the binomial expansion of the given equation ${{\left( d-5 \right)}^{6}}$, then we explain it with the help of Pascal’s triangle.
${{\left( d-5 \right)}^{6}}={{d}^{6}}-30{{d}^{5}}+375{{d}^{4}}-2500{{d}^{3}}+9375{{d}^{2}}-18750d+15625$.
Pascal’s triangle helps to find the coefficients for the expansion of the ${{\left( x+a \right)}^{n}}$, where n decides the number of times, we continue with the triangle expansion and the added value with x (for general case a) decides the multiplier. We multiply with ${{a}^{n}},a=0\left( 1 \right)n$ in a consecutive manner.
We first draw the triangle values till the ${{7}^{th}}$ row where it starts with 1 at the top.

Every coefficient is the addition of the previous two coefficients on its top. These coefficients are made for the expansion of the term ${{\left( x+1 \right)}^{n}}$. For particular we took the value of $n=6$ and that’s why we took 6 rows after the first value of 1 at the top.
Now instead of a, we have to multiply with -5 as for the equation ${{\left( d-5 \right)}^{6}}$ we have $a=-5$.
The relative coefficients are 1, 6, 15, 20, 15, 6 ,1. We multiply them with ${{\left( -5 \right)}^{0}},{{\left( -5 \right)}^{1}},{{\left( -5 \right)}^{2}},{{\left( -5 \right)}^{3}},{{\left( -5 \right)}^{4}},{{\left( -5 \right)}^{5}},{{\left( -5 \right)}^{6}}$ respectively.
Therefore, the actual coefficients are
$\begin{align}
  & {{\left( -5 \right)}^{0}}\times 1=1 \\
 & {{\left( -5 \right)}^{1}}\times 6=-30 \\
 & {{\left( -5 \right)}^{2}}\times 15=375 \\
 & {{\left( -5 \right)}^{3}}\times 20=-2500 \\
 & {{\left( -5 \right)}^{4}}\times 15=9375 \\
 & {{\left( -5 \right)}^{5}}\times 6=-18750 \\
 & {{\left( -5 \right)}^{6}}\times 1=15625 \\
\end{align}$

Therefore, the expansion is ${{\left( d-5 \right)}^{6}}={{d}^{6}}-30{{d}^{5}}+375{{d}^{4}}-2500{{d}^{3}}+9375{{d}^{2}}-18750d+15625$.

Note: In binomial expansion these coefficients are used in the form of combination where the expansion is ${{\left( x+a \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{a}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{a}^{1}}+...+{}^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}+...+{}^{n}{{C}_{n-1}}{{x}^{1}}{{a}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{0}}{{a}^{n}}$.
The general coefficient value for ${{\left( r+1 \right)}^{th}}$ term is ${}^{n}{{C}_{r}}$ where ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$.