How do i get component wihout x ? Thank you.

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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

# "the term in the expansion of" \ \ ( 1/2 x - 1/x )^4 \ "without" \ x \ \ "is:" \quad \ \ 3/2 \ . #

Explanation:

# "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 \ . #

Feb 21, 2018

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

Explanation:

Given expression # ( 1/2 x - 1/x )^4 #
Rewriting it as

# [( x/2 - 1/x )^2]^2 #
#=> ( bar(x^2/4 -1)+ 1/x^2 )^2 #
#=> ( x^2/4 -1)^2+2( x^2/4 -1) 1/x^2+1/x^4 #
# => x^4/16 -x^2/2+1+ 1/2 -2/x^2+1/x^4 #
# => x^4/16 -x^2/2+3/2 -2/x^2+1/x^4 #

We see that term without #x# is #3/2#