Question

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

Answer 1

Hint: To expand these types of expressions ${{\left(x-5 \right)}^{6}}$mostly we use binomial expansion and Pascal's triangle formula. By using these two formulas we can easily expand. The above given equation ${{\left(x-5 \right)}^{6}}$ is in the form of ${{\left (a+b \right)}^{n}}$, where let $a$ is equals to $x$ and $b$ is equals to $\left (-5 \right)$ and $n$ is equals to the power of the expression which is $6$. In mathematics, the binomial expansion describes the algebraic expansion of powers of a binomial. The expression of the binomial expansion is: ${{\left( a+b \right)}^{n}}{{=}^{n}}{{c}_{0}}{{a}^{n}}{{b}^{0}}{{+}^{n}}{{c}_{1}}{{a}^{n-1}}{{b}^{1}}+.................{{+}^{n}}{{c}_{n}}{{a}^{0}}{{b}^{n}}$ , where $^{n}{{c}_{0}}{{,}^{n}}{{c}_{1,}}..........{{,}^{n}}{{c}_{n}}$ are the combinations. the general formula of combinations is: ${{ C} _ {n, k}} =\dfrac {n!}{K! \left (n-k \right)!}$, where $n$ is the population and $k$ are the picks. By using the combination formula we can rewrite the binomial expansion as: $\Rightarrow {{\left (a+b \right)}^{n}}={{a}^{n}}+n{{a}^{n-1}}b+\dfrac{n\left (n-1 \right)}{2!}{{a} ^{n-2}}{{b}^{2}}+....................+{{b}^{n}}$ .

Complete step by step solution:
Now expanding the given expression ${{\left(x-5 \right)}^{6}}$by using the binomial expansion which is
 $\Rightarrow {{\left (a+b \right)}^{n}}={{a}^{n}}+n{{a}^{n-1}}b+\dfrac{n\left (n-1 \right)}{2!}{{a} ^{n-2}}{{b}^{2}}+....................+{{b}^{n}}$ then, we get
\[\begin{align}
  & \Rightarrow {{\left(x-5 \right)}^{6}}={{x}^{6}}-6{{x}^{5}}\left (5 \right)+\dfrac{30{{x}^{4}}{{\left (-5 \right)}^{2}}}{2!}+\dfrac {120{{x} ^ {3}} {{\left (-5 \right)}^{3}}}{3!}+\dfrac {360{{x} ^ {2}} {{\left (-5 \right)}^{4}}}{4!}+\dfrac {720x {{\left (-5 \right)}^{5}}+}{5!}\dfrac{720{{\left (-5 \right)}^{6}}}{\left (6! \right)} \\
 & \Rightarrow {{\left(x-5 \right)}^{6}}={{x}^{6}}-30{{x}^{5}}+375{{x}^{4}}-2500{{x}^{3}}+9375{{x}^{2}}-18750x+15625 \\
\end{align}\]
Hence by using the binomial expansion we get the expanded form of ${{\left(x-5 \right)}^{6}}$is\[{{\left (x-5 \right)}^{6}}={{x}^{6}}-30{{x}^{5}}+375{{x}^{4}}-2500{{x}^{3}}+9375{{x}^{2}}-18750x+15625\].
Now we will use Pascal's triangle formula to expand the given expression${{\left(x-5 \right)}^{6}}$. The Pascal's triangle is a never ending equilateral triangle of numbers which follow the rule of adding the two numbers to get the number below. The first row of Pascal's triangle is ${{\left (a+b \right)}^{0}}$. So for ${{\left(x-5 \right)}^{6}}$ we are looking for the 7^th row of the Pascal's triangle for coefficients:

Now by expanding the given expression ${{\left(x-5 \right)}^{6}}$by using Pascal's triangle formula we get,
$\begin {align}
  & \Rightarrow {{\left(x-5 \right)}^{6}}={{x}^{6}}-6\times {{x} ^ {5}}\times 5+15{{x} ^ {4}}\times 25-20{{x} ^ {3}}\times 125+15{{x} ^ {2}}\times 625-6x\times 3125+15625 \\
 & \\
\end{align}$\[\Rightarrow {{\left (x-5 \right)}^{6}}={{x}^{6}}-30{{x}^{5}}+375{{x}^{4}}-2500{{x}^{3}}+9375{{x}^{2}}-18750x+15625\]
Hence we get the expanded form of ${{\left(x-5 \right)}^{6}}$is same as we solved above which is \[{{\left (x-5 \right)}^{6}}={{x}^{6}}-30{{x}^{5}}+375{{x}^{4}}-2500{{x}^{3}}+9375{{x}^{2}}-18750x+15625\]

Note: We can go wrong in the calculation part, so use a calculator to solve these types of questions. We discussed two methods to solve these types of expression ${{\left(x-5 \right)}^{6}}$ one is binomial expansion and another is Pascal's triangle formula.