Please help. I'm not sure how to do this quickly without multiplying it all out?

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2 Answers
May 18, 2018

The answer to (i) is #240#.
The answer to (ii) is #200#.

Explanation:

We can do this by using Pascal's Triangle, shown below.

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(i)

Since the exponent is #6#, we need to use the sixth row in the triangle, which includes #color(purple)(1, 6, 15, 20, 15, 6)# and #color(purple)1#. Basically, we will use #color(blue)1# as the first term and #color(red)(2x)# as the second. Then, we can create the following equation. The exponent of the first term increases by #1# each time and the exponent of the second term decreases by #1# with each term from the triangle.

#(color(purple)1*color(blue)(1^0)*color(red)((2x)^6))+(color(purple)6*color(blue)(1^1)*color(red)((2x)^5))+(color(purple)15*color(blue)(1^2)*color(red)((2x)^4))+(color(purple)20*color(blue)(1^3)*color(red)((2x)^3))+(color(purple)15*color(blue)(1^4)*color(red)((2x)^2))+(color(purple)6*color(blue)(1^5)*color(red)((2x)^1))+(color(purple)1*color(blue)(1^6)*color(red)((2x)^0))#

Then, we can simplify it.

#64x^6+192x^5+240x^4+160x^3+60x^2+12x+1#

Therefore, the coefficient of #x^4# is #240#.

(ii)

We already know the expansion of #(1+2x)^6#. Now, we can multiply the two expressions together.

#color(brown)(1-x(1/4))*color(orange)(64x^6+192x^5+240x^4+160x^3+60x^2+12x+1)#

The coefficient of the #x# in #1-x(1/4)# is #1#. So, we know that it will raise the values of the exponents in the other expression by #1#. Because we need the coefficient of #x^4#, we just need to multiply #160x^3# by #1-x(1/4)#.

#160x^3-40x^4#

Now, we need to add it #240x^4#. This is one part of the solution of #240x^4*(1-x(1/4))#, due to the multiplication by #1#. It is significant because it also has an exponent of #4#.

#-40x^4+240x^4=200x^4#

Therefore, the coefficient is #200#.

May 18, 2018

i. #240x^4#
ii. #200x^4#

Explanation:

The binomial expansion for #(a+bx)^c# can be represented as:
#sum_(n=0)^c(c!)/(n!(c-n)!)a^(c-n)(bx)^n#

For part 1 we only need when #n=4#:
#(6!)/(4!(6-4)!)1^(6-4)(2x)^4#

#720/(24(2))16x^4#

#720/48 16x^4#

#15*16x^4#

#240x^4#

For part 2, we also need the #x^3# term because of the #x/4#

#(6!)/(3!(6-3)!)1^(6-3)(2x)^3#

#720/(3!(3)!)8x^3#

#720/(6^2)8x^3#

#720/36 8x^3#

#20*8x^3#

#160x^3#

#160x^3(-x/4)=-40x^4#

#-40x^4+240x^4=200x^4#