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

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

Explanation

Solution

We first write the binomial expansion of (d−5)6{{\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 (x+a)n{{\left( x+a \right)}^{n}}. Then we find the coefficients of the equation (d−5)6{{\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 (d−5)6{{\left( d-5 \right)}^{6}}, then we explain it with the help of Pascal’s triangle.
(d−5)6=d6−30d5+375d4−2500d3+9375d2−18750d+15625{{\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 (x+a)n{{\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 an,a=0(1)n{{a}^{n}},a=0\left( 1 \right)n in a consecutive manner.
We first draw the triangle values till the 7th{{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 (x+1)n{{\left( x+1 \right)}^{n}}. For particular we took the value of n=6n=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 (d−5)6{{\left( d-5 \right)}^{6}} we have a=−5a=-5.
The relative coefficients are 1, 6, 15, 20, 15, 6 ,1. We multiply them with (−5)0,(−5)1,(−5)2,(−5)3,(−5)4,(−5)5,(−5)6{{\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
(−5)0×1=1 (−5)1×6=−30 (−5)2×15=375 (−5)3×20=−2500 (−5)4×15=9375 (−5)5×6=−18750 (−5)6×1=15625 \begin{aligned} & {{\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{aligned}

Therefore, the expansion is (d−5)6=d6−30d5+375d4−2500d3+9375d2−18750d+15625{{\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 (x+a)n=nC0xna0+nC1xn−1a1+...+nCrxn−rar+...+nCn−1x1an−1+nCnx0an{{\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 (r+1)th{{\left( r+1 \right)}^{th}} term is nCr{}^{n}{{C}_{r}} where nCr=n!(n−r)!×r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}.