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avatar+216 

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

Can someone please help me i expanded it but I don't know which one is the constant term 

 

When I expanded it I got this \(46656x^6+15552x^4+2160x^2+160+\frac{20}{3x^2}+\frac{4}{27x^4}+\frac{1}{729x^6} \)

 Nov 7, 2021
 #1
avatar+159 
+2

Hello. The constant term is the term in which there is no variable. In your case, that is 160. There is no variable multiplying by it. 

 

Remember: Variables is an unknown value. 

 

and a constant is a known value with no variables. 

 Nov 7, 2021
 #2
avatar+115335 
+2

Hi JMaster,

 

I'm giving you one last chance to be polite.   (not just to me)

 

You should give yourself a point Mathygoo  wink

 

--------------------------------------

 

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

 

The general term is 

 

\(\begin{pmatrix} 6\\ r \end{pmatrix} \displaystyle \left(\frac{1}{3x}\right)^r (6x)^{6-r} \)

 

You need the term where the power of x is 0

 

so you need to solve this    \(\displaystyle \left(\frac{1}{x}\right)^r (x)^{6-r} =x^0\)

 

when you solve for r you can then sub that  r  value into the general term to get the constant that you require.

 Nov 7, 2021
edited by Melody  Nov 7, 2021
 #4
avatar+216 
+1

Thanks Melody

Jmaster10  Nov 7, 2021
 #3
avatar+216 
+1

Thanks guys it means a lot and that makes so much more sense

 Nov 7, 2021

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