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

 Nov 20, 2020
 #1
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By the Binomial Theorem, the constant term is 220.

 Nov 20, 2020
 #2
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Constant term   

 

C(6,3) (6x)^3  ( 1 / (3x) )^3  =

 

20 * (6)^3 * x^3  *  ( 1 / 27)  * ( 1/x^3)  =

 

20 * 6^3 / 27  =

 

20 * 216 / 27  =

 

20  *  8  = 

 

160

 

Verified here    :   https://www.wolframalpha.com/input/?i=%286x+%2B+1%2F+%283x%29+%29%5E6

 

 

cool cool cool 

 Nov 20, 2020
edited by CPhill  Nov 20, 2020

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