"One way to do this is to refer to the Binomial Theorem. "
"Recall the Binomial Theorem gives:"
\qquad \qquad \qquad \qquad \qquad \qquad ( a + b )^n \ = \ sum_{k=0}^{n} C(n, k) a^{n-k} b^k . \qquad \qquad \qquad \qquad \qquad \quad \ (1)
"In our case:" \qquad \qquad n=4, \qquad \qquad a =1/2 x, \qquad \qquad b = - 1/x. \
"So, substituting these into eqn. (1), we get:"
( 1/2 x - 1/x )^4 \ = \ sum_{k=0}^{4} C(4, k) ( 1/2 x )^{ 4-k } ( - 1/x )^k . \qquad \qquad \qquad (2)
"We want the term without" \ x. \ "Let's look carefully at the" \ \ k^{ mbox{th} } \ \
"term in the expansion in eqn. (2):"
\qquad \qquad \qquad \ k^{ mbox{th} } \ \ "term" \ \ = \ C(4, k) ( 1/2 x )^{ 4-k } ( -1/x)^k
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ C(4, k) ( 1/2 )^{ 4-k } x^{4-k} (-1)^k ( 1/x )^{ k }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ C(4, k) (-1)^k ( 1/2 )^{ 4-k } x^{ 4-k } x^{ -k }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ C(4, k) (-1)^k ( 1/2 )^{ 4-k } x^{ 4-k -k }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ C(4, k) (-1)^k ( 1/2 )^{ 4-k } x^{ 4-2k }.
"So, we have:"
\qquad \qquad \qquad \quad \ \ k^{ mbox{th} } \ \ "term" \ = \ C(4, k) (-1)^k ( 1/2 )^{ 4-k } x^{ 4-2k }. \qquad \qquad \qquad (3)
"We want the term without" \ \ x \ \ "in it. The idea here is to note"
"that this can also be thought of as the term where" \ \ x \ \ "has"
"exponent 0. Using eqn. (3), we see that the" \ \ k^{ mbox{th} } \ \ "term has" \ \ x
"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"
"that equation:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 4-2k = 0.
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 2k = 4.
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ k = 2.
"So, we conclude:"
\qquad \qquad "the second term in the expansion in eqn. (2) is the term"
\qquad \qquad \qquad \qquad "without" \ \ x.
"Now we calculate that term, using eqn. (3):"
\quad 2^{ mbox{nd} } \ "term in expansion in eqn. (2)"
\qquad \qquad \qquad \qquad \qquad \qquad \ = \ C(4, 2) (-1)^2 ( 1/2 )^{ 4-2} x^{ 4-2}
\qquad \qquad \qquad \qquad \qquad \qquad \ = \ { 4! } / { 2! 2! } cdot 1 cdot( 1/2 )^{ 2} x^{ 0}
\qquad \qquad \qquad \qquad \qquad \qquad \ = \ { 4 cdot 3 } / { 1 cdot 2 } cdot 1/2^{ 2} cdot 1
\qquad \qquad \qquad \qquad \qquad \qquad \ = \ { color{red}cancel{ 4 } cdot 3 } / { 1 cdot 2 } cdot 1/ color{red}cancel{ 2^{ 2} )
\qquad \qquad \qquad \qquad \qquad \qquad \ = \ { 3 } / { 1 cdot 2 }
\qquad \qquad \qquad \qquad \qquad \qquad \ = \ { 3 } / { 2 } \quad.
"This is our answer."
"Summarizing:"
"the term in the expansion of" \ \ ( 1/2 x - 1/x )^4 \ "without" \ x \ \ "is:" \quad \ \ 3/2 \ .
