# How do i get component wihout x ? Thank you.

## I know the n-k+1 and k-1 thingy.. but i have no idea if i should do something with it if there is 1/x and 1/2x. Thank you.

##### 2 Answers
Feb 21, 2018

$\text{the term in the expansion of" \ \ ( 1/2 x - 1/x )^4 \ "without" \ x \ \ "is:} \setminus \quad \setminus \setminus \frac{3}{2} \setminus .$

#### Explanation:

$\text{One way to do this is to refer to the Binomial Theorem. }$

$\text{Recall the Binomial Theorem gives:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {\left(a + b\right)}^{n} \setminus = \setminus {\sum}_{k = 0}^{n} C \left(n , k\right) {a}^{n - k} {b}^{k} . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \left(1\right)$

$\text{In our case:} \setminus q \quad \setminus q \quad n = 4 , \setminus q \quad \setminus q \quad a = \frac{1}{2} x , \setminus q \quad \setminus q \quad b = - \frac{1}{x} . \setminus$

$\text{So, substituting these into eqn. (1), we get:}$

${\left(\frac{1}{2} x - \frac{1}{x}\right)}^{4} \setminus = \setminus {\sum}_{k = 0}^{4} C \left(4 , k\right) {\left(\frac{1}{2} x\right)}^{4 - k} {\left(- \frac{1}{x}\right)}^{k} . \setminus q \quad \setminus q \quad \setminus q \quad \left(2\right)$

$\text{We want the term without" \ x. \ "Let's look carefully at the} \setminus \setminus {k}^{m b \otimes \left\{t h\right\}} \setminus \setminus$
$\text{term in the expansion in eqn. (2):}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus {k}^{m b \otimes \left\{t h\right\}} \setminus \setminus \text{term} \setminus \setminus = \setminus C \left(4 , k\right) {\left(\frac{1}{2} x\right)}^{4 - k} {\left(- \frac{1}{x}\right)}^{k}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus C \left(4 , k\right) {\left(\frac{1}{2}\right)}^{4 - k} {x}^{4 - k} {\left(- 1\right)}^{k} {\left(\frac{1}{x}\right)}^{k}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus C \left(4 , k\right) {\left(- 1\right)}^{k} {\left(\frac{1}{2}\right)}^{4 - k} {x}^{4 - k} {x}^{- k}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus C \left(4 , k\right) {\left(- 1\right)}^{k} {\left(\frac{1}{2}\right)}^{4 - k} {x}^{4 - k - k}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus C \left(4 , k\right) {\left(- 1\right)}^{k} {\left(\frac{1}{2}\right)}^{4 - k} {x}^{4 - 2 k} .$

$\text{So, we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus {k}^{m b \otimes \left\{t h\right\}} \setminus \setminus \text{term} \setminus = \setminus C \left(4 , k\right) {\left(- 1\right)}^{k} {\left(\frac{1}{2}\right)}^{4 - k} {x}^{4 - 2 k} . \setminus q \quad \setminus q \quad \setminus q \quad \left(3\right)$

$\text{We want the term without" \ \ x \ \ "in it. The idea here is to note}$
$\text{that this can also be thought of as the term where" \ \ x \ \ "has}$
$\text{exponent 0. Using eqn. (3), we see that the" \ \ k^{ mbox{th} } \ \ "term has} \setminus \setminus x$
$\text{with exponent:" \quad 4-2k. "So the term with zero exponent for" \ \ x \ "occurs where:" \ \ 4-2k = 0. "So to find that term, we must solve}$
$\text{that equation:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 4 - 2 k = 0.$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus 2 k = 4.$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus k = 2.$

$\text{So, we conclude:}$

$\setminus q \quad \setminus q \quad \text{the second term in the expansion in eqn. (2) is the term}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{without} \setminus \setminus x .$

$\text{Now we calculate that term, using eqn. (3):}$

$\setminus \quad {2}^{m b \otimes \left\{n d\right\}} \setminus \text{term in expansion in eqn. (2)}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus C \left(4 , 2\right) {\left(- 1\right)}^{2} {\left(\frac{1}{2}\right)}^{4 - 2} {x}^{4 - 2}$

 \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { 4! } / { 2! 2! } cdot 1 cdot( 1/2 )^{ 2} x^{ 0}

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus \frac{4 \cdot 3}{1 \cdot 2} \cdot \frac{1}{2} ^ \left\{2\right\} \cdot 1$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus \frac{\textcolor{red}{\cancel{4}} \cdot 3}{1 \cdot 2} \cdot \frac{1}{\textcolor{red}{\cancel{{2}^{2}}}}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus \frac{3}{1 \cdot 2}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus \frac{3}{2} \setminus \quad .$

$\text{This is our answer.}$

$\text{Summarizing:}$

$\text{the term in the expansion of" \ \ ( 1/2 x - 1/x )^4 \ "without" \ x \ \ "is:} \setminus \quad \setminus \setminus \frac{3}{2} \setminus .$

Feb 21, 2018

Alternate solution
Since power of the expression is $4$, this can be used.

#### Explanation:

Given expression ${\left(\frac{1}{2} x - \frac{1}{x}\right)}^{4}$
Rewriting it as

${\left[{\left(\frac{x}{2} - \frac{1}{x}\right)}^{2}\right]}^{2}$
$\implies {\left(\overline{{x}^{2} / 4 - 1} + \frac{1}{x} ^ 2\right)}^{2}$
$\implies {\left({x}^{2} / 4 - 1\right)}^{2} + 2 \left({x}^{2} / 4 - 1\right) \frac{1}{x} ^ 2 + \frac{1}{x} ^ 4$
$\implies {x}^{4} / 16 - {x}^{2} / 2 + 1 + \frac{1}{2} - \frac{2}{x} ^ 2 + \frac{1}{x} ^ 4$
$\implies {x}^{4} / 16 - {x}^{2} / 2 + \frac{3}{2} - \frac{2}{x} ^ 2 + \frac{1}{x} ^ 4$

We see that term without $x$ is $\frac{3}{2}$