The general term in the binomial expansion of $(a+b)^n$ is given by the formula: $$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Problem 14: Obtain the fourth term in the expansion of $(x-y)^9$. Here, $a=x$, $b=-y$, $n=9$. For the fourth term, $k+1=4$, so $k=3$. Step 1: Apply the general term formula. $$T_4 = \binom{9}{3} (x)^{9-3} (-y)^3$$ Step 2: Calculate the binomial coefficient. $$\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$$ Step 3: Substitute the values and simplify. $$T_4 = 84 x^6 (-y^3)$$ $$T_4 = -84x^6y^3$$ The correct option is B. The fourth term is $\boxed{\text{B. } -84x^6y^3}$. Problem 15: Obtain the fourth term in the expansion of $(x^2 + 3)^6$. Here, $a=x^2$, $b=3$, $n=6$. For the fourth term, $k+1=4$, so $k=3$. Step 1: Apply the general term formula. $$T_4 = \binom{6}{3} (x^2)^{6-3} (3)^3$$ Step 2: Calculate the binomial coefficient. $$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$ Step 3: Substitute the values and simplify. $$T_4 = 20 (x^2)^3 (3^3)$$ $$T_4 = 20 (x^6) (27)$$ $$T_4 = 540x^6$$ The correct option is B. The fourth term is $\boxed{\text{B. } 540x^6}$. Problem 16: Obtain the sixth term in the expansion of $(3x - 2y)^9$. Here, $a=3x$, $b=-2y$, $n=9$. For the sixth term, $k+1=6$, so $k=5$. Step 1: Apply the general term formula. $$T_6 = \binom{9}{5} (3x)^{9-5} (-2y)^5$$ Step 2: Calculate the binomial coefficient. $$\binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 9 \times 2 \times 7 = 126$$ Step 3: Substitute the values and simplify. $$T_6 = 126 (3x)^4 (-2y)^5$$ $$T_6 = 126 (3^4 x^4) ((-2)^5 y^5)$$ $$T_6 = 126 (81 x^4) (-32 y^5)$$ $$T_6 = 126 \times 81 \times (-32) x^4 y^5$$ $$T_6 = 10206 \times (-32) x^4 y^5$$ $$T_6 = -326592 x^4 y^5$$ The correct option is D. The sixth term is $\boxed{\text{D. } -326592x^4y^5}$. Problem 17: Obtain the second term in the expansion of $(2x - 1)^2$. Here, $a=2x$, $b=-1$, $n=2$. For the second term, $k+1=2$, so $k=1$. Step 1: Apply the general term formula. $$T_2 = \binom{2}{1} (2x)^{2-1} (-1)^1$$ Step 2: Calculate the binomial coefficient. $$\binom{2}{1} = \frac{2!}{1!(2-1)!} = \frac{2!}{1!1!} = 2$$ Step 3: Substitute the values and simplify. $$T_2 = 2 (2x)^1 (-1)^1$$ $$T_2 = 2 (2x) (-1)$$ $$T_2 = -4x$$ The correct option is C. The second term is $\boxed{\text{C. } -4x}$. Problem 18: What is the coefficient of $x^3$ in $(2 + 3x)^4$? Here, $a=2$, $b=3x$, $n=4$. We need to find the term where the power of $x$ is 3. The general term is $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. Step 1: Write the general term for this expansion. $$T_{k+1} = \binom{4}{k} (2)^{4-k} (3x)^k$$ $$T_{k+1} = \binom{4}{k} (2)^{4-k} (3^k x^k)$$ Step 2: Identify the value of $k$ for $x^3$. For the term containing $x^3$, we must have $x^k = x^3$, so $k=3$. Step 3: Substitute $k=3$ into the general term formula. $$T_{3+1} = T_4 = \binom{4}{3} (2)^{4-3} (3^3 x^3)$$ $$T_4 = \binom{4}{3} (2)^1 (3^3) x^3$$ Step 4: Calculate the binomial coefficient and simplify. $$\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = 4$$ $$T_4 = 4 \times 2 \times 27 x^3$$ $$T_4 = 8 \times 27 x^3$$ $$T_4 = 216 x^3$$ The coefficient of $x^3$ is 216. The correct option is D. The coefficient of $x^3$ is $\boxed{\text{D. } 216}$.