Question:

Step 1: Write down the general term of the binomial expansion. The general term, $T_{r+1}$, in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. For the given expansion $\left(x-\dfrac{3}{x^2}\right)^n$, we have $a=x$ and $b=-\dfrac{3}{x^2}$. $$T_{r+1} = \binom{n}{r} (x)^{n-r} \left(-\dfrac{3}{x^2}\right)^r$$ Simplifying the expression for $T_{r+1}$: $$T_{r+1} = \binom{n}{r} x^{n-r} (-3)^r (x^{-2})^r$$ $$T_{r+1} = \binom{n}{r} (-3)^r x^{n-r-2r}$$ $$T_{r+1} = \binom{n}{r} (-3)^r x^{n-3r}$$ Step 2: Determine the coefficients of the first three terms. The first three terms correspond to $r=0$, $r=1$, and $r=2$. For the first term ($r=0$): The coefficient is $\binom{n}{0} (-3)^0 = 1 \cdot 1 = 1$. For the second term ($r=1$): The coefficient is $\binom{n}{1} (-3)^1 = n \cdot (-3) = -3n$. For the third term ($r=2$): The coefficient is $\binom{n}{2} (-3)^2 = \dfrac{n(n-1)}{2} \cdot 9 = \dfrac{9n(n-1)}{2}$. Step 3: Use the sum of the coefficients to find the value of $n$. We are given that the sum of the coefficients of the first three terms is 376. $$1 + (-3n) + \dfrac{9n(n-1)}{2} = 376$$ $$1 - 3n + \dfrac{9n^2 - 9n}{2} = 376$$ To clear the denominator, multiply the entire equation by 2: $$2 - 6n + 9n^2 - 9n = 752$$ Rearrange into a standard quadratic equation form: $$9n^2 - 15n + 2 - 752 = 0$$ $$9n^2 - 15n - 750 = 0$$ Divide the equation by 3 to simplify: $$3n^2 - 5n - 250 = 0$$ Solve the quadratic equation for $n$ using the quadratic formula $n = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $$n = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-250)}}{2(3)}$$ $$n = \dfrac{5 \pm \sqrt{25 + 3000}}{6}$$ $$n = \dfrac{5 \pm \sqrt{3025}}{6}$$ Calculate the square root: $\sqrt{3025} = 55$. $$n = \dfrac{5 \pm 55}{6}$$ Two possible values for $n$: $n_1 = \dfrac{5 + 55}{6} = \dfrac{60}{6} = 10$ $n_2 = \dfrac{5 - 55}{6} = \dfrac{-50}{6} = -\dfrac{25}{3}$ Since $n \in \mathbb{N}$ (natural numbers), we must choose the positive integer value. Therefore, $n=10$. Step 4: Determine the value of $r$ for the coefficient of $x^4$. Substitute $n=10$ into the general term expression from Step 1: $$T_{r+1} = \binom{10}{r} (-3)^r x^{10-3r}$$ We want to find the coefficient of $x^4$. So, we set the exponent of $x$ equal to 4: $$10 - 3r = 4$$ $$3r = 10 - 4$$ $$3r = 6$$ $$r = 2$$ Step 5: Calculate the coefficient of $x^4$. Substitute $n=10$ and $r=2$ into the coefficient part of the general term: Coefficient of $x^4 = \binom{10}{2} (-3)^2$ Calculate the binomial coefficient: $$\binom{10}{2} = \dfrac{10!}{2!(10-2)!} = \dfrac{10 \times 9}{2 \times 1} = 45$$ Calculate the power of -3: $$(-3)^2 = 9$$ Multiply these values to find the coefficient of $x^4$: $$\text{Coefficient of } x^4 = 45 \times 9 = 405$$ The final answer is $\boxed{405}$.