# Binomila theorum-examzz help

1. Jun 18, 2006

### angel_eyez

i don't get this question

in the expansion of (3/x - x^3)^8 find

a) the constant term ( is dere an easier way to find this instead of expandin the equation n den finding it)

b) the term containin x^12

2. Jun 18, 2006

### Andrew Mason

The expansion of $(x + a)^n$ gives terms:

$$(x + a)^n = \sum_{k=0}^n \left(\begin{array}{c} n\\k\end{array}\right) x^ka^{n-k}$$

where the coefficient:

$$\left(\begin{array}{c} n\\k\end{array}\right) = \frac{n!}{(n-k)!k!}$$

So multiply your expression by -1/x^8 and then do the binomial expansion. Then multiply the binomial terms by -1/x^8 to find the constant term and one containing x^12 (if there is one).

$$(3/x - x^3)^8 = (\frac{-1}{x}(x^4-3))^8 = \frac{1}{x^8}\sum_{k=0}^n \left(\begin{array}{c} 8\\k\end{array}\right) x^{4k}(-3)^{8-k}$$

Can you do the rest from that?

AM

Last edited: Jun 18, 2006
3. Jun 18, 2006

### angel_eyez

i still kind adont get it..can u plz complete it

4. Jun 19, 2006

### HallsofIvy

Staff Emeritus
It's just doing the arithmetic now.
$$(3/x - x^3)^8 = (\frac{-1}{x}(x^4-3))^8 = \frac{1}{x^8}\sum_{k=0}^n \left(\begin{array}{c} 8\\k\end{array}\right) x^{4k}(-3)^{8-k}$$

The "kth" term is
$$(-3)^{8- k}\left(\begin{array}{c} 8\\k\end{array}\right)x^{-8}x^{4k}= (-3)^{8- k}\left(\begin{array}{c} 8\\k\end{array}\right)x^{4k-8}$$

a) The constant term has no x term or an x0 term. For what value of k is 4k- 8= 0? What is
$$(-3)^{8- k}\left(\begin{array}{c} 8\\k\end{array}\right)$$
for that k?
b) For what value of k is 4k- 8= 12? What is
$$(-3)^{8- k}\left(\begin{array}{c} 8\\k\end{array}\right)$$
for that k?