# Thread: Constant term

1. ## Constant term

How do I find the constant term in ((-x)^3+(-3/x))^24 ?

2. ## Binomial theorem.

Originally Posted by weasley74
How do I find the constant term in ((-x)^3+(-3/x))^24 ?

$\displaystyle (a+b)^n=\sum_{k=0}^{n}\binom{n}{k}(a)^{n-k}b^{k}$

for yours we get

$\displaystyle ([(-x)]^3+(-3x^{-1}))^{24}=\sum_{k=0}^{24}\binom{24}{k}(-x^{3})^{24-k}(-3x^{-1})^{k}$

This should get you started. I have to go

3. I've tried using this, but I get the wrong answer in the end..

4. You need the term such that:
(Ax^3)^a*(Bx^-1)^(24-a) is constant.

This is only satisfied when a = 6

Hence the constant term is:

(-1)^6*(-3)^18*(24 choose 6)

5. I'm back here is how you finish.

$\displaystyle ([(-x)]^3+(-3x^{-1}))^{24}=\sum_{k=0}^{24}\binom{24}{k}(-x^{3})^{24-k}(-3x^{-1})^{k}$

$\displaystyle \sum_{k=0}^{24} \binom{24}{k}(-1)^{24-k}x^{72-3k}(-3)^{k}(x)^{-k}= \sum_{k=0}^{24} \binom{24}{k}(3)^{k}x^{72-4k}$

Sinc ewe want the constant term we want the exponent on the x to be zero.

solving the equation

$\displaystyle 72-4k=0 \iff k=18$

we can generate the constant term using k=18

$\displaystyle \binom{24}{18}(-x^3)^{24-18}(-3x^{-1})^{18}= \binom{24}{18}(-x^3)^6\left(\frac{-3}{x} \right)^{18}=\binom{24}{18}(3)^{18}=52125248137444$

6. thanks! got it now! =)