Expand (x−2y)⁵ using the binomial theorem (a+b)ⁿ = sum_k=0ⁿ nk aⁿ−k b^k with a=x, b=−2y, and n=5.

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1. Find the binomial expansion of $x^2(x-2y)^5$ .

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 $n=5$ .
$(x-2y)^5 = \binom{5}{0} x^5 (-2y)^0 + \binom{5}{1} x^4 (-2y)^1 + \binom{5}{2} x^3 (-2y)^2 + \binom{5}{3} x^2 (-2y)^3 + \binom{5}{4} x^1 (-2y)^4 + \binom{5}{5} x^0 (-2y)^5$ $= 1 \cdot x^5 \cdot 1 + 5 \cdot x^4 \cdot (-2y) + 10 \cdot x^3 \cdot (4y^2) + 10 \cdot x^2 \cdot (-8y^3) + 5 \cdot x \cdot (16y^4) + 1 \cdot 1 \cdot (-32y^5)$ $= x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5$
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 $\left(x - \frac{3}{x}\right)^4$ .
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 $n=4$ .
$T_{k+1} = \binom{4}{k} x^{4-k} (-3x^{-1})^k$ $T_{k+1} = \binom{4}{k} x^{4-k} (-3)^k (x^{-1})^k$ $T_{k+1} = \binom{4}{k} (-3)^k x^{4-k-k}$ $T_{k+1} = \binom{4}{k} (-3)^k x^{4-2k}$
For the term independent of $x$ , the power of $x$ must be 0.
$4-2k = 0$ $2k = 4$ $k = 2$
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 $(1-y)^{\frac{1}{2}}$ .
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 $((x+1)(2x-\frac{1}{x})^3)$ .
Expand $(2x-\frac{1}{x})^3$ using the binomial theorem.
Let $a=2x$ and $b=-\frac{1}{x} = -x^{-1}$ , $n=3$ .
$(2x-x^{-1})^3 = \binom{3}{0}(2x)^3(-x^{-1})^0 + \binom{3}{1}(2x)^2(-x^{-1})^1 + \binom{3}{2}(2x)^1(-x^{-1})^2 + \binom{3}{3}(2x)^0(-x^{-1})^3$ $= 1 \cdot (8x^3) \cdot 1 + 3 \cdot (4x^2) \cdot (-x^{-1}) + 3 \cdot (2x) \cdot (x^{-2}) + 1 \cdot 1 \cdot (-x^{-3})$ $= 8x^3 - 12x^{2-1} + 6x^{1-2} - x^{-3}$ $= 8x^3 - 12x + 6x^{-1} - x^{-3}$
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:
$\boxed{\text{ } 6 \text{}}$

Expand (x-2y)⁵ using the binomial theorem (a+b)ⁿ = sum_k=0ⁿ nk aⁿ-k b^k with a=x, b=-2y, and n=5.

Step-by-step solution

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 $n=5$ .

Multiply the expansion by $x^2$ .

Write the general term $T_{k+1}$ for the binomial expansion $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ .

For the term independent of $x$ , the power of $x$ must be 0.

Substitute $k=2$ into the general term formula.

Use the binomial expansion for $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$ .

Expand $(2x-\frac{1}{x})^3$ using the binomial theorem.

Multiply the expanded form by $(x+1)$ .

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More Mathematics Questions

Expand (x-2y)⁵ using the binomial theorem (a+b)ⁿ = sum_k=0ⁿ nk aⁿ-k b^k with a=x, b=-2y, and n=5.

Step-by-step solution

Expand (x−2y)5(x-2y)^5(x−2y)5 using the binomial theorem (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k(a+b)n=∑k=0n​(kn​)an−kbk with a=xa=xa=x, b=−2yb=-2yb=−2y, and n=5n=5n=5.

Multiply the expansion by x2x^2x2.

Write the general term Tk+1T_{k+1}Tk+1​ for the binomial expansion (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k(a+b)n=∑k=0n​(kn​)an−kbk.

For the term independent of xxx, the power of xxx must be 0.

Substitute k=2k=2k=2 into the general term formula.

Use the binomial expansion for (1+x)n=1+nx+n(n−1)2!x2+…(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots(1+x)n=1+nx+2!n(n−1)​x2+….

Expand (2x−1x)3(2x-\frac{1}{x})^3(2x−x1​)3 using the binomial theorem.

Multiply the expanded form by (x+1)(x+1)(x+1).

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More Mathematics Questions

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 $n=5$ .

Multiply the expansion by $x^2$ .

Write the general term $T_{k+1}$ for the binomial expansion $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ .

For the term independent of $x$ , the power of $x$ must be 0.

Substitute $k=2$ into the general term formula.

Use the binomial expansion for $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$ .

Expand $(2x-\frac{1}{x})^3$ using the binomial theorem.

Multiply the expanded form by $(x+1)$ .