1. **State the problem:** We are given the first three terms of the binomial expansion of $ (1 + kx)^{16} $ as $1$, $-4x$, and $px^2$, where $k$ and $p$ are constants. (a)(i) Find $k$. (a)(ii) Find $p$. (b) Given $g(x) = \left(2 + \frac{16}{x}\right)(1 + kx)^{16}$ and the value of $k$ from part (a), find the coefficient of the $x^2$ term in the expansion of $g(x)$. --- 2. **Recall the binomial expansion formula:** $$ (1 + kx)^{16} = \sum_{r=0}^{16} \binom{16}{r} (1)^{16-r} (kx)^r = \sum_{r=0}^{16} \binom{16}{r} k^r x^r $$ The first three terms correspond to $r=0,1,2$: - Term 0: $\binom{16}{0} k^0 x^0 = 1$ - Term 1: $\binom{16}{1} k x = 16 k x$ - Term 2: $\binom{16}{2} k^2 x^2 = 120 k^2 x^2$ 3. **Use given terms to find $k$ and $p$:** - Given term 1 is $-4x$, so: $$ 16 k x = -4 x $$ Cancel $x$: $$ 16 \cancel{x} k = -4 \cancel{x} $$ $$ 16 k = -4 $$ $$ k = \frac{-4}{16} = -\frac{1}{4} $$ - Given term 2 is $p x^2$, so: $$ 120 k^2 x^2 = p x^2 $$ Cancel $x^2$: $$ 120 k^2 = p $$ Substitute $k = -\frac{1}{4}$: $$ p = 120 \left(-\frac{1}{4}\right)^2 = 120 \times \frac{1}{16} = \frac{120}{16} = \frac{15}{2} = 7.5 $$ 4. **Find the $x^2$ term in $g(x) = \left(2 + \frac{16}{x}\right)(1 + kx)^{16}$:** Expand $g(x)$: $$ g(x) = 2(1 + kx)^{16} + \frac{16}{x}(1 + kx)^{16} $$ We want the coefficient of $x^2$ in $g(x)$. - From $2(1 + kx)^{16}$, the $x^2$ term is: $$ 2 \times p x^2 = 2 p x^2 $$ - From $\frac{16}{x}(1 + kx)^{16}$, multiply $16/x$ by each term of $(1 + kx)^{16}$: The $x^2$ term in $g(x)$ from this part comes from the $x^3$ term in $(1 + kx)^{16}$ multiplied by $16/x$: The $x^3$ term in $(1 + kx)^{16}$ is: $$ \binom{16}{3} k^3 x^3 = 560 k^3 x^3 $$ Multiply by $16/x$: $$ 16/x \times 560 k^3 x^3 = 16 \times 560 k^3 x^{3-1} = 16 \times 560 k^3 x^2 $$ So the $x^2$ term from this part is: $$ 16 \times 560 k^3 x^2 = 8960 k^3 x^2 $$ 5. **Sum the $x^2$ terms from both parts:** $$ x^2 \text{ term in } g(x) = 2 p x^2 + 8960 k^3 x^2 = \left(2 p + 8960 k^3\right) x^2 $$ Substitute $k = -\frac{1}{4}$ and $p = \frac{15}{2}$: Calculate $k^3$: $$ k^3 = \left(-\frac{1}{4}\right)^3 = -\frac{1}{64} $$ Calculate $2p$: $$ 2 p = 2 \times \frac{15}{2} = 15 $$ Calculate $8960 k^3$: $$ 8960 \times \left(-\frac{1}{64}\right) = -140 $$ Sum: $$ 15 + (-140) = -125 $$ **Final answer:** The coefficient of $x^2$ in $g(x)$ is $-125$. --- **Summary:** - (a)(i) $k = -\frac{1}{4}$ - (a)(ii) $p = \frac{15}{2}$ - (b) Coefficient of $x^2$ in $g(x)$ is $-125$