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What is the constant term of the expansion of (6x + 1/(3x))^8?

 Jul 19, 2021
 #1
avatar+68 
+2

Alright, let's break this down by simplifying it.

 

\((6x+1/3x)^2 \)

 

\(=(6x+\frac{1}{3x})^2\)

 

Now we can apply the formula

 

 \((a+b)^2=a^2+2ab+b^2\)

 

or in this case,

 

\((6x + \frac{1}{3x})^2 = 6x^2+2(6x)\frac{1}{3x}+\frac{1}{3x}^2\)

 

which will become:

 

6x^2 + 4 + 1/3x2, since 2*6x is 12x, and 12x+1/3x is 12x/3x. Divide by 3x and you get the answer, \(\fbox{4}\)

 Jul 19, 2021
 #2
avatar+2250 
0

Welcome to web2.0calc. :)

Nice solution, but it says raised to the 8th instead of 2nd. 

To get the constant term, we need 4 "6x" and 4 "1/(3x)". 

Using the binomial theorem, the coefficient is 8C4 = 70.

70 * (6x)^4 * (1/(3x))^4 = 1120

 

=^._.^=

catmg  Jul 19, 2021
edited by catmg  Jul 20, 2021
 #3
avatar+68 
+1

Ah. I constantly forget to pay attention to the details. Thanks for the heads up!

PBJcatalinasandwich  Jul 19, 2021

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