## May 18, 2018

The answer to (i) is $240$.
The answer to (ii) is $200$.

#### Explanation:

We can do this by using Pascal's Triangle, shown below. (i)

Since the exponent is $6$, we need to use the sixth row in the triangle, which includes $\textcolor{p u r p \le}{1 , 6 , 15 , 20 , 15 , 6}$ and $\textcolor{p u r p \le}{1}$. Basically, we will use $\textcolor{b l u e}{1}$ as the first term and $\textcolor{red}{2 x}$ as the second. Then, we can create the following equation. The exponent of the first term increases by $1$ each time and the exponent of the second term decreases by $1$ with each term from the triangle.

$\left(\textcolor{p u r p \le}{1} \cdot \textcolor{b l u e}{{1}^{0}} \cdot \textcolor{red}{{\left(2 x\right)}^{6}}\right) + \left(\textcolor{p u r p \le}{6} \cdot \textcolor{b l u e}{{1}^{1}} \cdot \textcolor{red}{{\left(2 x\right)}^{5}}\right) + \left(\textcolor{p u r p \le}{15} \cdot \textcolor{b l u e}{{1}^{2}} \cdot \textcolor{red}{{\left(2 x\right)}^{4}}\right) + \left(\textcolor{p u r p \le}{20} \cdot \textcolor{b l u e}{{1}^{3}} \cdot \textcolor{red}{{\left(2 x\right)}^{3}}\right) + \left(\textcolor{p u r p \le}{15} \cdot \textcolor{b l u e}{{1}^{4}} \cdot \textcolor{red}{{\left(2 x\right)}^{2}}\right) + \left(\textcolor{p u r p \le}{6} \cdot \textcolor{b l u e}{{1}^{5}} \cdot \textcolor{red}{{\left(2 x\right)}^{1}}\right) + \left(\textcolor{p u r p \le}{1} \cdot \textcolor{b l u e}{{1}^{6}} \cdot \textcolor{red}{{\left(2 x\right)}^{0}}\right)$

Then, we can simplify it.

$64 {x}^{6} + 192 {x}^{5} + 240 {x}^{4} + 160 {x}^{3} + 60 {x}^{2} + 12 x + 1$

Therefore, the coefficient of ${x}^{4}$ is $240$.

(ii)

We already know the expansion of ${\left(1 + 2 x\right)}^{6}$. Now, we can multiply the two expressions together.

$\textcolor{b r o w n}{1 - x \left(\frac{1}{4}\right)} \cdot \textcolor{\mathmr{and} a n \ge}{64 {x}^{6} + 192 {x}^{5} + 240 {x}^{4} + 160 {x}^{3} + 60 {x}^{2} + 12 x + 1}$

The coefficient of the $x$ in $1 - x \left(\frac{1}{4}\right)$ is $1$. So, we know that it will raise the values of the exponents in the other expression by $1$. Because we need the coefficient of ${x}^{4}$, we just need to multiply $160 {x}^{3}$ by $1 - x \left(\frac{1}{4}\right)$.

$160 {x}^{3} - 40 {x}^{4}$

Now, we need to add it $240 {x}^{4}$. This is one part of the solution of $240 {x}^{4} \cdot \left(1 - x \left(\frac{1}{4}\right)\right)$, due to the multiplication by $1$. It is significant because it also has an exponent of $4$.

$- 40 {x}^{4} + 240 {x}^{4} = 200 {x}^{4}$

Therefore, the coefficient is $200$.

May 18, 2018

i. $240 {x}^{4}$
ii. $200 {x}^{4}$

#### Explanation:

The binomial expansion for ${\left(a + b x\right)}^{c}$ can be represented as:
sum_(n=0)^c(c!)/(n!(c-n)!)a^(c-n)(bx)^n

For part 1 we only need when $n = 4$:
(6!)/(4!(6-4)!)1^(6-4)(2x)^4

$\frac{720}{24 \left(2\right)} 16 {x}^{4}$

$\frac{720}{48} 16 {x}^{4}$

$15 \cdot 16 {x}^{4}$

$240 {x}^{4}$

For part 2, we also need the ${x}^{3}$ term because of the $\frac{x}{4}$

(6!)/(3!(6-3)!)1^(6-3)(2x)^3

720/(3!(3)!)8x^3

$\frac{720}{{6}^{2}} 8 {x}^{3}$

$\frac{720}{36} 8 {x}^{3}$

$20 \cdot 8 {x}^{3}$

$160 {x}^{3}$

$160 {x}^{3} \left(- \frac{x}{4}\right) = - 40 {x}^{4}$

$- 40 {x}^{4} + 240 {x}^{4} = 200 {x}^{4}$