Find the constant term of the following expansion:

\[(x-(1/5x ^{2}))^{9}\]

I would guess it was supposed to be written as: \[ \left( x - \frac{1}{5x^2} \right)^9 \] Since that would lead to a better result, I believe. Just need to know for sure that it's correct. :D

yeah it's supposed to look like that

I didn't see the rule that disallowed it

If we think about it intuitively, how do we get a constant term if we have a rational expression? We'll have some power of x in the numerator and some power of x in the denominator, so we'd want those x's to essentially cancel out, right? This occurs when the power of x in the numerator is equal to the power of x in the denominator, because then division cancels them. So, we want to focus on \(when\) that happens. Well, if we think about how the powers would line up in the expansion... x^9 / (x^0)^2 = x^9 / x^0 x^8 / (x^2)^1 = x^8 / x^2 x^7 / (x^2)^2 = x^7 / x^4 x^6 / (x^2)^3 = x^6 / x^6 <--- Here! x^5 / (x^2)^4 = x^5 / x^8 x^4 / (x^2)^5 = x^4 / x^10 .. we can see that's not going to give us anymore possible constant terms other than x^6 / x^6, the fourth term in the expansion. Now, we just need to find the coefficient of this term. The binomial expansion multiplies "9C3" on the outside of the fourth term, and then we have the extra "1/5^3" because the "1/(5x^2)" is cubed, and we took out the (1/x^2)^3 considering the cancellation earlier.

x^9 / (x^2)^0 = x^9 / x^0 *** Not sure what I was writing there. lol

and it's negative then because we have a negative term to an odd power, so we can see that this leads to -84/125.

So basically, we're just taking the bigger problem and using a bit of logic about how the math works out to break it apart into an easier problem. If it makes sense, it is quicker this way to just find the value of interest to us, the term that is constant when the two variables cancel, rather than expand it all out.

i'm confused as to what just happened.

Uhh, Hero posted a link for the expansion of your expression leading to the answer, and I just tried to explain another way to find it, without having to necessarily expand it but using the idea of expansions.

ok.

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