In the binomial expansion of $\left(x + \frac{k}{x}\right)^5$, where $k$ is a positive integer, the coefficients of $\frac{1}{x^3}$ and $\frac{1}{x}$ are the same. Find the value of $k$. Without expanding all the terms, explain why there is no constant term in the expansion of $\left(1 - 3x^2\right)\left(x + \frac{k}{x}\right)^5$. Use your graph to estimate value of $p$ and of $k$, the temperature of the liquid after 35 minutes, the number of minutes it takes for the temperature of the liquid to drop by 75% of its initial value. the total distance travelled during the first 4 seconds.

1. Write down the general term of the binomial expansion

The general term $T_{r+1}$ in the expansion of $\left(x + \frac{k}{x}\right)^5$ is:
$T_{r+1} = \binom{5}{r} x^{5-r} \left(\frac{k}{x}\right)^r$

2. Simplify the general term to isolate the power of x

$T_{r+1} = \binom{5}{r} k^r x^{5-r} x^{-r} = \binom{5}{r} k^r x^{5-2r}$

3. Find the value of r for the term in 1/x^3

For the term in $\frac{1}{x^3}$ (or $x^{-3}$):
$5 - 2r = -3$
$2r = 8 \implies r = 4$
The coefficient is $\binom{5}{4} k^4 = 5k^4$.

4. Find the value of r for the term in 1/x

For the term in $\frac{1}{x}$ (or $x^{-1}$):
$5 - 2r = -1$
$2r = 6 \implies r = 3$
The coefficient is $\binom{5}{3} k^3 = 10k^3$.

5. Equate the coefficients and solve for k

Since the coefficients are the same:
$5k^4 = 10k^3$
Since $k$ is a positive integer, $k \neq 0$, so we can divide by $5k^3$:
$k = 2$

1. Find the general term for the expansion of $(x + \frac{k}{x})^5$.

The general term is given by:
$T_{r+1} = \binom{5}{r} x^{5-r} \left(\frac{k}{x}\right)^r = \binom{5}{r} k^r x^{5-2r}$

2. Identify the powers of $x$ required from the binomial expansion to form a constant term.

To obtain a constant term (i.e., $x^0$) in the expansion of $\left(1 - 3x^2\right)\left(x + \frac{k}{x}\right)^5$, we need to multiply:

• $1$ by the $x^0$ term from $\left(x + \frac{k}{x}\right)^5$
• $-3x^2$ by the $x^{-2}$ term from $\left(x + \frac{k}{x}\right)^5$

3. Show that these required powers of $x$ do not exist in the binomial expansion.

For the $x^0$ term: $5 - 2r = 0 \implies r = 2.5$
For the $x^{-2}$ term: $5 - 2r = -2 \implies r = 3.5$
Since $r$ must be an integer ($0 \leqslant r \leqslant 5$), neither the $x^0$ term nor the $x^{-2}$ term exists in the expansion of $\left(x + \frac{k}{x}\right)^5$. Therefore, there is no constant term in the overall expansion.