Another one Princess — let's solve it. Question 27: The term independent of $(x)$ in the expansion of $(1-2x)^2$ is Step 1: Expand the expression $(1-2x)^2$. $$(1-2x)^2 = (1)^2 - 2(1)(2x) + (2x)^2$$ $$= 1 - 4x + 4x^2$$ Step 2: Identify the term independent of $x$. The term independent of $x$ is the constant term. In the expansion $1 - 4x + 4x^2$, the constant term is $1$. The term independent of $x$ is $\boxed{\text{1}}$. Since $1$ is not among options a) 15, b) -15, c) 32, d) -32, the correct option is e) None of the above. Question 28: In the expansion of $(a-b)^n$, the sum of the coefficients is Step 1: To find the sum of coefficients of a polynomial, substitute $1$ for each variable. For the expansion of $(a-b)^n$, substitute $a=1$ and $b=1$. Sum of coefficients $= (1-1)^n = 0^n$. Step 2: Evaluate $0^n$. If $n > 0$, then $0^n = 0$. In binomial expansion, $n$ is typically a positive integer. The sum of the coefficients is $\boxed{\text{0}}$. This matches option a) 0. Question 29: The middle term in the expansion of $(1+x)^{10}$ is Step 1: Determine the number of terms in the expansion. For $(a+b)^n$, there are $n+1$ terms. Here $n=10$, so there are $10+1=11$ terms. Step 2: Find the position of the middle term. Since there are $11$ terms (an odd number), there is one middle term. Its position is $\frac{n+1+1}{2} = \frac{11+1}{2} = \frac{12}{2} = 6$. So, it's the $6$-th term. Step 3: Use the general term formula $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. For the $6$-th term, $k+1=6 \implies k=5$. For $(1+x)^{10}$, $a=1$, $b=x$, $n=10$. $$T_6 = \binom{10}{5} (1)^{10-5} (x)^5$$ $$T_6 = \binom{10}{5} (1)^5 (x)^5$$ $$T_6 = \binom{10}{5} x^5$$ The question asks for "the middle term", and the options are powers of $x$. This implies we should identify the variable part of the middle term. The middle term is $\boxed{\binom{10}{5} x^5}$. The variable part is $x^5$. This matches option a) $x^5$. Question 30: The coefficient of the term containing $x^3$ in the expansion of $(1-2x)^{-4}$ 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 $n = -4$. We need the term containing $x^3$, which corresponds to the term with $y^3$. The coefficient of $y^3$ is $\frac{n(n-1)(n-2)}{3!}$. Step 2: Substitute $n=-4$ and $y=-2x$ into the $y^3$ term. The term is $\frac{(-4)(-4-1)(-4-2)}{3!} (-2x)^3$. $$= \frac{(-4)(-5)(-6)}{3 \times 2 \times 1} (-2)^3 x^3$$ $$= \frac{-120}{6} (-8) x^3$$ $$= (-20) (-8) x^3$$ $$= 160x^3$$ Step 3: Identify the coefficient of $x^3$. The coefficient of $x^3$ is $160$. The coefficient of the term containing $x^3$ is $\boxed{\text{160}}$. Since $160$ is not among options a) -4, b) -8, c) 8, d) 4, the correct option is e) None of the above. Question 31: What is the term independent of $x$ in the expansion of $(1+x)^7$? Step 1: Use the general term formula $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. For $(1+x)^7$, $a=1$, $b=x$, $n=7$. The term independent of $x$ is the term where the power of $x$ is $0$. So, $k=0$. Step 2: Substitute $k=0$ into the general term formula. $$T_{0+1} = \binom{7}{0} (1)^{7-0} (x)^0$$ $$T_1 = \binom{7}{0} (1)^7 (1)$$ Step 3: Calculate the value. $$\binom{7}{0} = 1$$ $$T_1 = 1 \cdot 1 \cdot 1 = 1$$ The term independent of $x$ is $\boxed{\text{1}}$. This matches option a) 1. Question 32: The coefficient of the second term in the expansion of $(3-2x)^5$ is Step 1: Use the general term formula $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. For $(3-2x)^5$, $a=3$, $b=-2x$, $n=5$. We need the second term, so $k+1=2 \implies k=1$. Step 2: Substitute $k=1$ into the general term formula. $$T_{1+1} = \binom{5}{1} (3)^{5-1} (-2x)^1$$ $$T_2 = \binom{5}{1} (3)^4 (-2x)$$ Step 3: Calculate the values. $$\binom{5}{1} = 5$$ $$3^4 = 81$$ $$T_2 = 5 \cdot 81 \cdot (-2x)$$ $$T_2 = 405 \cdot (-2x)$$ $$T_2 = -810x$$ Step 4: Identify the coefficient of the second term. The coefficient of the second term is $-810$. The coefficient of the second term is $\boxed{\text{-810}}$. Since $-810$ is not among options a) -120, b) 240, c) -240, d) 120, the correct option is e) None of the above. Question 33: The sum of the coefficients in the expansion of $(a+b)^n$ is Step 1: To find the sum of coefficients of a polynomial, substitute $1$ for each variable. For the expansion of $(a+b)^n$, 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) $2^n$. Send me the next one 📸