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What is the constant term in the expansion of \(\left(6x+\dfrac{1}{3x}\right)^6\)?

 Jan 7, 2020
 #1
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0

There is no constant.  Do you mean coefficient?

 Jan 7, 2020
 #2
avatar+106519 
+2

(6x   +  1 / (3x) )^6

 

The constant term is

 

C(6,3)  (6x)3 ( 1 / (3x) )3  =

 

20 * (63) (x3) ( 1 / (33)) ( 1 / x3)  =  

 

20 * 216 / 27  =  

 

20 * 8  =  

 

160

 

 

cool cool cool

 Jan 7, 2020
 #3
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0

You lost me at C(6,3).

Guest Jan 7, 2020
 #4
avatar+106519 
+2

OK......here's another way to solve this

 

C(6, n) (6x)n (  1 / (3x) ) 6 - n        and we can write

 

C (6, n)  ( 6)n (x)n  ( 1 / 3)6-n ( 1 / x) 6-n

 

C (6, n)  (6)n( 1/3)6-n  ( x)n  ( 1/x)6-n

 

C(6, n) (6)n (1/3)6-n ( x)n  ( x-1)6-n

 

C(6, n)  (6)n (1/3)6-n (x)n ( x)n - 6

 

C(6, n) (6)n(1/3)6-n ( x) 2n - 6

 

We need to   have    2n - 6  = 0    to eliminate the variable, x

 

Then   n  must    = 3

 

So  we have

 

C(6, 3) (6)3 (1/3)6-3   

 

20 * 216 * (1/3)3

 

20 * 216 *  (1/27)  =

 

20 * 8  =

 

160

 

cool cool cool

 Jan 7, 2020

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