Here are the solutions to the problems: 1. Find the binomial expansion of $x^2(x-2y)^5$. Step 1: Expand $(x-2y)^5$ using the binomial theorem $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ with $a=x$, $b=-2y$, and $n=5$. $$ (x-2y)^5 = \binom{5}{0} x^5 (-2y)^0 + \binom{5}{1} x^4 (-2y)^1 + \binom{5}{2} x^3 (-2y)^2 + \binom{5}{3} x^2 (-2y)^3 + \binom{5}{4} x^1 (-2y)^4 + \binom{5}{5} x^0 (-2y)^5 $$ $$ = 1 \cdot x^5 \cdot 1 + 5 \cdot x^4 \cdot (-2y) + 10 \cdot x^3 \cdot (4y^2) + 10 \cdot x^2 \cdot (-8y^3) + 5 \cdot x \cdot (16y^4) + 1 \cdot 1 \cdot (-32y^5) $$ $$ = x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5 $$ Step 2: Multiply the expansion by $x^2$. $$ x^2(x-2y)^5 = x^2(x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5) $$ $$ = x^7 - 10x^6y + 40x^5y^2 - 80x^4y^3 + 80x^3y^4 - 32x^2y^5 $$ The binomial expansion is: $$ \boxed{\text{ } x^7 - 10x^6y + 40x^5y^2 - 80x^4y^3 + 80x^3y^4 - 32x^2y^5 \text{}} $$ 2. Find the term independent of $x$ in the expansion of $\left(x - \frac{3}{x}\right)^4$. Step 1: Write the general term $T_{k+1}$ for the binomial expansion $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. Here, $a=x$, $b=-\frac{3}{x} = -3x^{-1}$, and $n=4$. $$ T_{k+1} = \binom{4}{k} x^{4-k} (-3x^{-1})^k $$ $$ T_{k+1} = \binom{4}{k} x^{4-k} (-3)^k (x^{-1})^k $$ $$ T_{k+1} = \binom{4}{k} (-3)^k x^{4-k-k} $$ $$ T_{k+1} = \binom{4}{k} (-3)^k x^{4-2k} $$ Step 2: For the term independent of $x$, the power of $x$ must be 0. $$ 4-2k = 0 $$ $$ 2k = 4 $$ $$ k = 2 $$ Step 3: Substitute $k=2$ into the general term formula. $$ T_{2+1} = T_3 = \binom{4}{2} (-3)^2 x^{4-2(2)} $$ $$ T_3 = \frac{4 \times 3}{2 \times 1} \cdot (9) \cdot x^0 $$ $$ T_3 = 6 \cdot 9 \cdot 1 $$ $$ T_3 = 54 $$ The term independent of $x$ is: $$ \boxed{\text{ } 54 \text{}} $$ 3. Find the first three terms in ascending powers of $y$ in the expansion of $(1-y)^{\frac{1}{2}}$. Step 1: Use the binomial expansion for $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$. Here, $x=-y$ and $n=\frac{1}{2}$. First term: $1$ Second term: $nx$ $$ \frac{1}{2}(-y) = -\frac{1}{2}y $$ Third term: $\frac{n(n-1)}{2!}x^2$ $$ \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!} (-y)^2 = \frac{\frac{1}{2}(-\frac{1}{2})}{2} (y^2) = \frac{-\frac{1}{4}}{2} y^2 = -\frac{1}{8}y^2 $$ The first three terms are: $$ \boxed{\text{ } 1 - \frac{1}{2}y - \frac{1}{8}y^2 \text{}} $$ 4. Find the coefficient of $x^{-1}$ in the expansion of $((x+1)(2x-\frac{1}{x})^3)$. Step 1: Expand $(2x-\frac{1}{x})^3$ using the binomial theorem. Let $a=2x$ and $b=-\frac{1}{x} = -x^{-1}$, $n=3$. $$ (2x-x^{-1})^3 = \binom{3}{0}(2x)^3(-x^{-1})^0 + \binom{3}{1}(2x)^2(-x^{-1})^1 + \binom{3}{2}(2x)^1(-x^{-1})^2 + \binom{3}{3}(2x)^0(-x^{-1})^3 $$ $$ = 1 \cdot (8x^3) \cdot 1 + 3 \cdot (4x^2) \cdot (-x^{-1}) + 3 \cdot (2x) \cdot (x^{-2}) + 1 \cdot 1 \cdot (-x^{-3}) $$ $$ = 8x^3 - 12x^{2-1} + 6x^{1-2} - x^{-3} $$ $$ = 8x^3 - 12x + 6x^{-1} - x^{-3} $$ Step 2: Multiply the expanded form by $(x+1)$. $$ (x+1)(8x^3 - 12x + 6x^{-1} - x^{-3}) $$ To find the coefficient of $x^{-1}$, we look for terms that multiply to give $x^{-1}$: • $x$ multiplied by a term with $x^{-2}$: There is no $x^{-2}$ term in $(8x^3 - 12x + 6x^{-1} - x^{-3})$. So, $x \cdot (0) = 0$. • $1$ multiplied by a term with $x^{-1}$: $1 \cdot (6x^{-1}) = 6x^{-1}$. The only term contributing to $x^{-1}$ is $6x^{-1}$. The coefficient of $x^{-1}$ is: $$ \boxed{\text{ } 6 \text{}} $$