Alright Princess, let's tackle these. Question 33: The sum of the coefficients in the expansion of $(a+b)^n$ is Step 1: To find the sum of coefficients, substitute $1$ for each variable in the expression. For $(a+b)^n$, substitute $a=1$ and $b=1$. Sum of coefficients $= (1+1)^n = 2^n$. The sum of the coefficients is $\boxed{2^n}$. This matches option c) $2^n$. Question 34: In the expansion of $(1+2x)^6$, the coefficient of $x^2$ is Step 1: Use the general term formula for binomial expansion: $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. Here, $a=1$, $b=2x$, and $n=6$. We want the coefficient of $x^2$, so we set $k=2$. Step 2: Substitute $k=2$ into the formula. $$T_{2+1} = \binom{6}{2} (1)^{6-2} (2x)^2$$ $$T_3 = \binom{6}{2} (1)^4 (2x)^2$$ Step 3: Calculate the binomial coefficient and simplify. $$\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$$ $$T_3 = 15 \cdot 1 \cdot (4x^2)$$ $$T_3 = 60x^2$$ The coefficient of $x^2$ is $60$. The coefficient of $x^2$ is $\boxed{\text{60}}$. This matches option b) 60. Question 35: The coefficient of $x^4$ in the expansion of $(1-x)^6$ is Step 1: Use the general term formula: $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. Here, $a=1$, $b=-x$, and $n=6$. We want the coefficient of $x^4$, so we set $k=4$. Step 2: Substitute $k=4$ into the formula. $$T_{4+1} = \binom{6}{4} (1)^{6-4} (-x)^4$$ $$T_5 = \binom{6}{4} (1)^2 (-x)^4$$ Step 3: Calculate the binomial coefficient and simplify. $$\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$$ $$T_5 = 15 \cdot 1 \cdot x^4$$ $$T_5 = 15x^4$$ The coefficient of $x^4$ is $15$. The coefficient of $x^4$ is $\boxed{\text{15}}$. This matches option c) 15. Question 36: In the expansion of $(2+3x)^4$, what is the coefficient of $x^2$? Step 1: Use the general term formula: $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. Here, $a=2$, $b=3x$, and $n=4$. We want the coefficient of $x^2$, so we set $k=2$. Step 2: Substitute $k=2$ into the formula. $$T_{2+1} = \binom{4}{2} (2)^{4-2} (3x)^2$$ $$T_3 = \binom{4}{2} (2)^2 (3x)^2$$ Step 3: Calculate the binomial coefficient and simplify. $$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$$ $$T_3 = 6 \cdot 4 \cdot (9x^2)$$ $$T_3 = 6 \cdot 4 \cdot 9 \cdot x^2$$ $$T_3 = 216x^2$$ The coefficient of $x^2$ is $216$. The coefficient of $x^2$ is $\boxed{\text{216}}$. Since $216$ is not among options a) -72, b) -72, c) 36, d) -36, the correct option is e) None of the above. Question 37: The coefficient of the term containing $x^2$ in the expansion of $(1+2x)^{-3}$ is Step 1: Use the generalized binomial theorem for $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots$ Here, $y=2x$ and $n=-3$. We need the term containing $x^2$, which corresponds to the term with $y^2$. Step 2: Substitute $n=-3$ and $y=2x$ into the $y^2$ term. The term is $\frac{n(n-1)}{2!}y^2$. $$= \frac{(-3)(-3-1)}{2!} (2x)^2$$ $$= \frac{(-3)(-4)}{2 \times 1} (4x^2)$$ $$= \frac{12}{2} (4x^2)$$ $$= 6 (4x^2)$$ $$= 24x^2$$ The coefficient of $x^2$ is $24$. The coefficient of the term containing $x^2$ is $\boxed{\text{24}}$. Since $24$ is not among options a) -3, b) -6, c) 6, d) 3, the correct option is e) None of the above. Question 38: The term independent of $x$ in the expansion of $(1-4x)^5$ is Step 1: Use the general term formula: $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. Here, $a=1$, $b=-4x$, and $n=5$. The term independent of $x$ is the term where the power of $x$ is $0$. This occurs when $k=0$. Step 2: Substitute $k=0$ into the formula. $$T_{0+1} = \binom{5}{0} (1)^{5-0} (-4x)^0$$ $$T_1 = \binom{5}{0} (1)^5 (1)$$ Step 3: Calculate the value. $$\binom{5}{0} = 1$$ $$T_1 = 1 \cdot 1 \cdot 1 = 1$$ The term independent of $x$ is $1$. The term independent of $x$ is $\boxed{\text{1}}$. Since $1$ is not among options a) -40, b) 40, c) -10, d) 10, the correct option is e) None of the above. What's next?