Write down, and simplify the terms in the expansion of $(2 + x)^5$ in ascending powers of $x$. Hence find the coefficient of $x^3$ in the expansion of $(2 + 3x + x^2)^5$. Using the result in part (ii), determine the coefficient of $x^{3}$ in the expansion of $\left(1+\dfrac{3}{2}x+\dfrac{1}{2}x^{2}\right)^{5}$. Calculate the value of the constant term in the expansion of $\left(1-\dfrac{1}{x}\right)\left(2+3x+x^{2}\right)^{5}$.

1. Write the binomial expansion structure for power 5.

$(2+x)^5=2^5+5(2^4)(x)+10(2^3)(x^2)+10(2^2)(x^3)+5(2)(x^4)+x^5$

2. Simplify the numerical coefficients and present in ascending powers of $x$.

$2^5=32,\quad 5(2^4)=80,\quad 10(2^3)=80,\quad 10(2^2)=40,\quad 5(2)=10$

So,
$(2+x)^5=32+80x+80x^2+40x^3+10x^4+x^5$

1. Substitute $3x+x^2$ for $x$ in the expansion from part (i).

From part (i),
$(2+x)^5=32+80x+80x^2+40x^3+10x^4+x^5.$
Let $x$ be replaced by $3x+x^2$:
$(2+3x+x^2)^5=32+80(3x+x^2)+80(3x+x^2)^2+40(3x+x^2)^3+ ext{higher powers}.$

2. Keep only the parts that can produce an $x^3$ term.

Since $3x+x^2$ has lowest power $x$, the terms $10(3x+x^2)^4$ and $(3x+x^2)^5$ have powers at least $x^4$, so they do not contribute to the coefficient of $x^3$.

Now find the $x^3$ contributions:
$80(3x+x^2)^2=80(9x^2+6x^3+x^4),$
so the $x^3$ contribution is
$80(6x^3)=480x^3.$
Also,
$40(3x+x^2)^3.$
For the $x^3$ term, take $(3x)^3=27x^3$, so the contribution is
$40(27x^3)=1080x^3.$

3. Add the $x^3$ contributions to obtain the coefficient.

Total $x^3$ term:
$480x^3+1080x^3=1560x^3.$
Therefore, the coefficient of $x^3$ is
$1560.$

1. Express the bracket as a constant multiple of the bracket used in part (ii).

$1+\frac{3}{2}x+\frac{1}{2}x^2=\frac{1}{2}(2+3x+x^2)$
Hence,
$\left(1+\frac{3}{2}x+\frac{1}{2}x^2\right)^5=\left[\frac{1}{2}(2+3x+x^2)\right]^5=\frac{1}{32}(2+3x+x^2)^5.$

2. Use the coefficient from part (ii) and multiply by the scale factor.

From part (ii), the coefficient of $x^3$ in $(2+3x+x^2)^5$ is $1200$.

Therefore, the coefficient of $x^3$ in $\frac{1}{32}(2+3x+x^2)^5$ is
$\frac{1}{32}\times 1200=37.5=\frac{75}{2}.$

1. Identify which terms from $(2+3x+x^2)^5$ contribute to the constant term after multiplication by $1-\dfrac{1}{x}$.

Let
$F(x)=(2+3x+x^2)^5.$
Then
$\left(1-\frac{1}{x}\right)F(x)=F(x)-\frac{F(x)}{x}.$
So the constant term required is
$(\text{constant term of }F)- (\text{coefficient of }x\text{ in }F).$

2. Find the constant term and coefficient of $x$ in $(2+3x+x^2)^5$, then subtract.

From
$(2+3x+x^2)^5=(2+(3x+x^2))^5,$
the constant term is
$2^5=32.$
Using the expansion from part (i),
$(2+u)^5=32+80u+80u^2+40u^3+10u^4+u^5,$
where $u=3x+x^2$.

The coefficient of $x$ comes only from
$80(3x+x^2),$
so the coefficient of $x$ in $F(x)$ is
$80\times 3=240.$

Hence the required constant term is
$32-240=-208.$