# Thread: Need help with a Pascal's Triangle question

1. ## Need help with a Pascal's Triangle question

($\displaystyle \frac{1}{2x^2}$ - 4x^5)^5

I know I need to use Pascal's Triangle, but the numbers and exponents in this particular one are overwhelming me and I'm having trouble. Any help would be appreciated. Also, sorry about the half Latex half "lazy" equation, it kept giving me errors on the back half and I couldn't figure out what was going on.

2. ## Re: Need help with a Pascal's Triangle question

($\displaystyle \frac{1}{2x^2}$ - 4x^5)^5

I know I need to use Pascal's Triangle, but the numbers and exponents in this particular one are overwhelming me and I'm having trouble. Any help would be appreciated. Also, sorry about the half Latex half "lazy" equation, it kept giving me errors on the back half and I couldn't figure out what was going on.
It's just an application of the Binomial expansion \displaystyle \displaystyle \begin{align*} (a + b)^n = \sum_{r = 0}^n{{n\choose{r}}a^{n-r}b^r} \end{align*}

It might get a little messy but just keep working through it.

3. ## Re: Need help with a Pascal's Triangle question

Working it through I got

(1/32X^10) + (-5/4X^3) + (20x^4) + (-160x^11) + (640x^18) + (-1024x^25)

Now I am 95% sure some or all of this is wrong. Am I on the right track at least?

4. ## Re: Need help with a Pascal's Triangle question

Working it through I got

(1/32X^10) + (-5/4X^3) + (20x^4) + (-160x^11) + (640x^18) + (-1024x^25)

Now I am 95% sure some or all of this is wrong. Am I on the right track at least?
Don't skip steps, if you are learning something new, do every step.

\displaystyle \displaystyle \begin{align*} (a + b)^n &= \sum_{r = 0}^n{{n\choose{r}}a^{n-r}b^r} \\ \left(\frac{1}{2x^2} - 4x^5\right)^5 &= \sum_{r = 0}^5{{5\choose{r}}\left(\frac{1}{2x^2}\right)^{5 - r}\left(-4x^5\right)^r} \\ &= {5\choose{1}}\left(\frac{1}{2x^2}\right)^5\left(-4x^5\right)^0 + {5\choose{1}}\left(\frac{1}{2x^2}\right)^4\left(-4x^5\right)^1 + {5\choose{2}}\left(\frac{1}{2x^2}\right)^3\left(-4x^5\right)^2 + {5\choose{3}}\left(\frac{1}{2x^2}\right)^2\left(-4x^5\right)^3 + {5\choose{4}}\left(\frac{1}{2x^2}\right)^1\left(-4x^5\right)^4 + {5\choose{5}}\left(\frac{1}{2x^2}\right)^0\left(-4x^5\right)^5 \end{align*}

Now simplify.

5. ## Re: Need help with a Pascal's Triangle question

$\displaystyle \text{Expand: }\:\left(\frac{1}{2x^2} - 4x^5\right)^5$
Your answer was: .$\displaystyle \frac{1}{32x^{10}} - \frac{5}{4x^3} + 20x^4 - 160x^{11} + 640x^{18} -1024x^{25}$