# Help! Binomial Theorem

1. Jul 2, 2005

### ms. confused

Find the term containing $$x^{20}$$ in $$(2x - x^4)^{14}$$.

I went $$t_{k+1}= _{14}C_{k}(2x)^{14-k}(-x^{4})^{k}$$

$$= 2x^{14-k}(-x^{4k})$$

First of all, am I on the right track? If so what exactly do I do from there?

2. Jul 2, 2005

### AKG

Something is definitely wrong. First of all, your binomial coefficient has disappeard on the second line. Secondly, you seem to pull the 2 out of (2x)14-k in an improper way. You also can't bring the k inside the bracket like that, since it's not ((-x)4)k, it's (-x[sup4[/sup])k. Do you see the difference? The exponent 4 applies only to the x, not to (-x), but the exponent k applies to the whole (-x4). You should end up with:

$$t_{k+1} = {{14}\choose k}(-2)^{14-k}x^{14 + 3k}$$

See if you can figure out why the above is right. Now, to find the term containing x20, find the value(s) of k that satisfy 14 + 3k = 20. The only value for k is 2. So plug in 2 for k, you'll get:

(91)(212)x20

3. Jul 2, 2005

### ms. confused

Okay I think I get your point about the $$-x^4$$ thing. But how did you get the 2x to turn into -2 and how did you get the 14+3k exponent?

4. Jul 2, 2005

### lurflurf

You want the x^20 term so if the exponent of x in the kth term is
4k+(14-k)=20 what is k?

5. Jul 3, 2005

### ms. confused

+3k... right...thanks I see it now!

6. Jul 3, 2005

### manjish

Take (x^14) common and find the third term in the expansion of (2-(x^3))^14 i.e.
14C2*(2^12)*(-1)^2.