Find the term independent of x in the expansion of (x − 3/x)4.

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Here are the solutions for problems 2, 3, and 4.

: Find the term independent of $x$ in the expansion of $\left(x - \frac{3}{x}\right)^4$ .

Write the general term of the binomial expansion.
The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ .
In this case, $a=x$ , $b=-\frac{3}{x} = -3x^{-1}$ , and $n=4$ .
$T_{r+1} = \binom{4}{r} (x)^{4-r} (-3x^{-1})^r$
Simplify the powers of $x$ .
$T_{r+1} = \binom{4}{r} x^{4-r} (-3)^r (x^{-1})^r$ $T_{r+1} = \binom{4}{r} (-3)^r x^{4-r} x^{-r}$ $T_{r+1} = \binom{4}{r} (-3)^r x^{4-2r}$
Find the value of $r$ for the term independent of $x$ .
For the term to be independent of $x$ , the exponent of $x$ must be 0.
$4 - 2r = 0$ $2r = 4$ $r = 2$
Substitute $r=2$ into the general term to find the term.
$T_{2+1} = T_3 = \binom{4}{2} (-3)^2 x^{4-2(2)}$ $T_3 = \frac{4!}{2!(4-2)!} (-3)^2 x^0$ $T_3 = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} (9) \times 1$ $T_3 = \frac{24}{4} (9)$ $T_3 = 6 \times 9$ $T_3 = 54$
The term independent of $x$ is $\boxed{54}$ .

: Find the first three terms in ascending powers of $y$ in the expansion of $(1-y)^{\frac{1}{2}}$ .
Use the binomial expansion formula for fractional exponents.
The formula for $(1+x)^n$ is $1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$ .
Here, $x = -y$ and $n = \frac{1}{2}$ .
Calculate the first term.
The first term is 1.
Calculate the second term.
$nx = \frac{1}{2}(-y) = -\frac{1}{2}y$
Calculate the third term.
$\frac{n(n-1)}{2!}x^2 = \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(-y)^2$ $= \frac{\frac{1}{2}(-\frac{1}{2})}{2 \times 1} (y^2)$ $= \frac{-\frac{1}{4}}{2} y^2$ $= -\frac{1}{8}y^2$
Combine the first three terms.
The first three terms in ascending powers of $y$ are $1 - \frac{1}{2}y - \frac{1}{8}y^2$ .
The final answer is $\boxed{1 - \frac{1}{2}y - \frac{1}{8}y^2}$ .

: Find the coefficient of $x^{-1}$ in the expansion of $((x+1)(2x-\frac{1}{x}))$ .
Expand the product of the two factors.
$(x+1)\left(2x-\frac{1}{x}\right) = x(2x) + x\left(-\frac{1}{x}\right) + 1(2x) + 1\left(-\frac{1}{x}\right)$ $= 2x^2 - 1 + 2x - \frac{1}{x}$
Rewrite the expression with negative exponents.
$= 2x^2 + 2x - 1 - x^{-1}$
Identify the term containing $x^{-1}$ and its coefficient.
The term containing $x^{-1}$ is $-x^{-1}$ .
The coefficient of $x^{-1}$ is $-1$ .
The final answer is $\boxed{-1}$ .

Find the term independent of x in the expansion of (x - (3)/(x))4.

Step-by-step solution

Write the general term of the binomial expansion.

Simplify the powers of $x$ .

Find the value of $r$ for the term independent of $x$ .

Substitute $r=2$ into the general term to find the term.

Use the binomial expansion formula for fractional exponents.

Calculate the first term.

Calculate the second term.

Calculate the third term.

Combine the first three terms.

Expand the product of the two factors.

Rewrite the expression with negative exponents.

Identify the term containing $x^{-1}$ and its coefficient.

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