# SOLVEDDetermine the constant term in.

#### Petrus

Determine the constant term in.

idk how to solve it

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#### skeeter

MHF Helper
each term in a binomial expansion is of the form ...

$$\displaystyle \binom{n}{k} a^{n-k} b^k$$

In this case, $$\displaystyle n = 21$$ ... the value of $$\displaystyle k$$ will be such that the variable factors cancel.

$$\displaystyle \binom{21}{18} (3x^6)^3\left(\frac{5}{x}\right)^{18}$$

#### Petrus

i did not get it, i dont think i really understand what they want me to answer, with other words i maybe dont understand problem... im confused

#### skeeter

MHF Helper
i did not get it, i dont think i really understand what they want me to answer, with other words i maybe dont understand problem... im confused
1. have you learned the binomial theorem?

2. do you understand that a constant term has no variable factors?

#### Petrus

1. have you learned the binomial theorem?

2. do you understand that a constant term has no variable factors?
hi!
We have not come here yet, but I want to study in advance. I did read but did not understand clearly i dont like my book

MHF Helper
1 person

#### Petrus

ok i did watch and i kinda understand but idk how in my problem to get out the constant term :S

#### skeeter

MHF Helper
I showed you how in post #2.

#### Salahuddin559

A constant terms comes out when a^n-k * b^k is a constant, and does not involve any variables. In your case, the binomial expansion has only one variable, x. That means, in (3x^6)^n - k * (5/x)^k does not contain any x. For it to happen, when n = 21, you need to equate the power of x to zero, since x^0 = 1, is a constant, and other powers yield terms with x, and so, not a constant term.

(x^6)^21 - k * (1/x)^k = x^0
(x^6)^21 - k * (x^-1)^k = x^0
(x^6)^21 - k * x^-k = x^0
x^(6(21-k) - k) = x^0

Equating coefficients, you get k = 18, which is what he used to get the constant term above.

Salahuddin
Maths online

#### Petrus

I solved this problem forgot to say it ty anyway