1. **State the problem:** We are given the first three terms in descending powers of $x$ in the expansion of $$\left( 2x^{2} - \frac{1}{4x} \right)^n$$ as $$256x^{16} + ax^{13} + bx^{c}$$ and need to find the integers $n, a, b,$ and $c$. 2. **Recall the binomial expansion formula:** The general term in the expansion of $(A - B)^n$ is $$T_{k+1} = \binom{n}{k} A^{n-k} (-B)^k$$ for $k=0,1,2,...,n$. 3. **Identify $A$ and $B$:** Here, $A = 2x^2$ and $B = \frac{1}{4x}$. 4. **Write the first three terms:** \begin{align*} T_1 &= (2x^2)^n = 2^n x^{2n} \\ T_2 &= \binom{n}{1} (2x^2)^{n-1} \left(-\frac{1}{4x}\right) = n \cdot 2^{n-1} x^{2(n-1)} \cdot \left(-\frac{1}{4x}\right) = -\frac{n}{4} 2^{n-1} x^{2n-2-1} = -\frac{n}{4} 2^{n-1} x^{2n-3} \\ T_3 &= \binom{n}{2} (2x^2)^{n-2} \left(-\frac{1}{4x}\right)^2 = \frac{n(n-1)}{2} 2^{n-2} x^{2(n-2)} \cdot \frac{1}{16 x^2} = \frac{n(n-1)}{2} 2^{n-2} \frac{1}{16} x^{2n-4-2} = \frac{n(n-1)}{2} \frac{2^{n-2}}{16} x^{2n-6} \end{align*} 5. **Simplify coefficients:** \begin{align*} T_1 &= 2^n x^{2n} \\ T_2 &= -\frac{n}{4} 2^{n-1} x^{2n-3} = -n 2^{n-3} x^{2n-3} \\ T_3 &= \frac{n(n-1)}{2} \frac{2^{n-2}}{16} x^{2n-6} = \frac{n(n-1)}{2} 2^{n-6} x^{2n-6} \end{align*} 6. **Match the first term to $256x^{16}$:** $$2^n x^{2n} = 256 x^{16}$$ So, powers of $x$ give: $$2n = 16 \implies n = 8$$ And coefficients: $$2^n = 256 \implies 2^8 = 256$$ This confirms $n=8$. 7. **Find $a$ from the second term:** $$a x^{13} = T_2 = -n 2^{n-3} x^{2n-3}$$ Substitute $n=8$: $$a x^{13} = -8 \cdot 2^{5} x^{13} = -8 \cdot 32 x^{13} = -256 x^{13}$$ So, $a = -256$. 8. **Find $b$ and $c$ from the third term:** $$b x^{c} = T_3 = \frac{n(n-1)}{2} 2^{n-6} x^{2n-6}$$ Substitute $n=8$: $$b x^{c} = \frac{8 \cdot 7}{2} 2^{2} x^{10} = 28 \cdot 4 x^{10} = 112 x^{10}$$ So, $b = 112$ and $c = 10$. **Final answers:** $$n=8, \quad a = -256, \quad b = 112, \quad c = 10$$