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Question: How do you expand \({\left( {x - 5} \right)^6}\) using Pascal’s triangle?...

How do you expand (x5)6{\left( {x - 5} \right)^6} using Pascal’s triangle?

Explanation

Solution

We first write the binomial expansion of (x5)6{\left( {x - 5} \right)^6}. Then we explain Pascal's triangle and the use 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 (x5)6{\left( {x - 5} \right)^6} using Pascal’s triangle.

Complete step-by-step answer:
First, we write down the binomial expansion of the given equation (x5)6{\left( {x - 5} \right)^6}, then we explain it with the help of Pascal’s triangle.
(x5)6=x630x5+375x42500x3+9375x218750x+15625\Rightarrow {\left( {x - 5} \right)^6} = {x^6} - 30{x^5} + 375{x^4} - 2500{x^3} + 9375{x^2} - 18750x + 15625
Pascal’s triangle helps to find the coefficients for the expansion of the (x5)6{\left( {x - 5} \right)^6}, 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^n}, a=0(1)na = 0\left( 1 \right)n consecutively.
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 (x5)6{\left( {x - 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,(5)1×6,(5)2×15,(5)3×20,(5)4×15,(5)5×6,(5)6×1\Rightarrow {\left( { - 5} \right)^0} \times 1,{\left( { - 5} \right)^1} \times 6,{\left( { - 5} \right)^2} \times 15,{\left( { - 5} \right)^3} \times 20,{\left( { - 5} \right)^4} \times 15,{\left( { - 5} \right)^5} \times 6,{\left( { - 5} \right)^6} \times 1
Simplify the terms,
1,30,375,2500,9375,18750,15625\Rightarrow 1, - 30,375, - 2500,9375, - 18750,15625

Hence, the expansion of (x5)6{\left( {x - 5} \right)^6} is x630x5+375x42500x3+9375x218750x+15625{x^6} - 30{x^5} + 375{x^4} - 2500{x^3} + 9375{x^2} - 18750x + 15625.

Note:
In binomial expansion, these coefficients are used in the form of combination where the expansion is
(x+a)n=nC0xna0+nC1xn1a1++nCrxnrar++nCn1x1an1+nCnx0an{\left( {x + a} \right)^n} = {}^n{C_0}{x^n}{a^0} + {}^n{C_1}{x^{n - 1}}{a^1} + \ldots + {}^n{C_r}{x^{n - r}}{a^r} + \ldots + {}^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 nCn=n!r!(nr)!{}^n{C_n} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}.