1. **Stating the problem:** Expand $e^{-12x}$ in ascending powers of $x$ as far as the term in $x^4$. Find the constant $n$ such that the coefficient of $x^4$ in the expansion of $e^{-12x}(1 - 6x^2)^n$ is $-936$. 2. **Formula and rules:** The exponential function expansion is given by: $$e^u = \sum_{r=0}^\infty \frac{u^r}{r!}$$ where $u = -12x$. The binomial expansion for $(1 + v)^n$ is: $$ (1 + v)^n = \sum_{r=0}^n \binom{n}{r} v^r $$ where $v = -6x^2$. 3. **Expand $e^{-12x}$ up to $x^4$ term:** $$e^{-12x} = 1 + (-12x) + \frac{(-12x)^2}{2!} + \frac{(-12x)^3}{3!} + \frac{(-12x)^4}{4!} + \cdots$$ Calculate each term: - $1$ - $-12x$ - $\frac{144x^2}{2} = 72x^2$ - $\frac{-1728x^3}{6} = -288x^3$ - $\frac{20736x^4}{24} = 864x^4$ So, $$e^{-12x} = 1 - 12x + 72x^2 - 288x^3 + 864x^4 + \cdots$$ 4. **Expand $(1 - 6x^2)^n$ up to $x^4$ term:** $$(1 - 6x^2)^n = \sum_{r=0}^n \binom{n}{r} (-6x^2)^r = \sum_{r=0}^n \binom{n}{r} (-6)^r x^{2r}$$ Terms up to $x^4$ correspond to $r=0,1,2$: - $r=0$: $1$ - $r=1$: $\binom{n}{1} (-6) x^2 = -6n x^2$ - $r=2$: $\binom{n}{2} 36 x^4 = 36 \frac{n(n-1)}{2} x^4 = 18 n(n-1) x^4$ So, $$(1 - 6x^2)^n = 1 - 6n x^2 + 18 n(n-1) x^4 + \cdots$$ 5. **Multiply expansions and find coefficient of $x^4$:** $$e^{-12x}(1 - 6x^2)^n = \left(1 - 12x + 72x^2 - 288x^3 + 864x^4\right) \times \left(1 - 6n x^2 + 18 n(n-1) x^4\right)$$ Collect terms up to $x^4$: - Constant term: $1 \times 1 = 1$ - $x$ term: $-12x \times 1 = -12x$ - $x^2$ term: $72x^2 \times 1 + 1 \times (-6n x^2) = (72 - 6n) x^2$ - $x^3$ term: $-288x^3 \times 1 + (-12x) \times (-6n x^2) = -288x^3 + 72n x^3 = (-288 + 72n) x^3$ - $x^4$ term: $864x^4 \times 1 + 72x^2 \times (-6n x^2) + 1 \times 18 n(n-1) x^4$ Calculate $x^4$ coefficient: $$864 - 432 n + 18 n(n-1) = 864 - 432 n + 18 n^2 - 18 n = 864 - 450 n + 18 n^2$$ 6. **Set coefficient of $x^4$ equal to $-936$ and solve for $n$:** $$18 n^2 - 450 n + 864 = -936$$ Add $936$ to both sides: $$18 n^2 - 450 n + 864 + 936 = 0$$ $$18 n^2 - 450 n + 1800 = 0$$ Divide entire equation by 18: $$\cancel{18} n^2 - \cancel{18} \times 25 n + \cancel{18} \times 100 = 0 \Rightarrow n^2 - 25 n + 100 = 0$$ 7. **Solve quadratic equation:** $$n = \frac{25 \pm \sqrt{25^2 - 4 \times 1 \times 100}}{2} = \frac{25 \pm \sqrt{625 - 400}}{2} = \frac{25 \pm \sqrt{225}}{2} = \frac{25 \pm 15}{2}$$ Two solutions: - $n = \frac{25 + 15}{2} = \frac{40}{2} = 20$ - $n = \frac{25 - 15}{2} = \frac{10}{2} = 5$ **Final answers:** - Expansion of $e^{-12x}$ up to $x^4$ term is: $$1 - 12x + 72x^2 - 288x^3 + 864x^4 + \cdots$$ - The constant $n$ such that the coefficient of $x^4$ in $e^{-12x}(1 - 6x^2)^n$ is $-936$ is $n = 5$ or $n = 20$.