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 \(46656x^6+15552x^4+2160x^2+160+\frac{20}{3x^2}+\frac{4}{27x^4}+\frac{1}{729x^6} \)

Jmaster10 Nov 7, 2021

#1**+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.

MathyGoo13 Nov 7, 2021

#2**+2 **

Hi JMaster,

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

You should give yourself a point Mathygoo

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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.

Melody Nov 7, 2021