# Math Help - Binomial expansions help please!

1. ## Binomial expansions help please!

please write out the expansions for each it would help me greatly understand how its done...

find the co efficient of x^6 in the expansion (1+3x)^12

find the term in x^3 in the expansion of (2+x)^4

find the binomial expansion of (2-x)^4 writing each term as simply as possible

2. Pascal's triangle is probably your best bet.
Pascal's triangle - Wikipedia, the free encyclopedia

Or if you know binomial expansion, you can just use that as well, which is included in the website above.

3. Originally Posted by coyoteflare
please write out the expansions for each it would help me greatly understand how its done...

find the co efficient of x^6 in the expansion (1+3x)^12
recall $(a + b)^n = \sum_{k = 0}^n {n \choose k} a^{n - k}b^k$

what does k have to be to get $x^6$?

find the term in x^3 in the expansion of (2+x)^4
use the same formula above, what does k have to be to get $x^3$

find the binomial expansion of (2-x)^4 writing each term as simply as possible

just expand based on the same formula again. you are writing out all the terms now

Originally Posted by terr13
Pascal's triangle is probably your best bet.
Pascal's triangle - Wikipedia, the free encyclopedia

Or if you know binomial expansion, you can just use that as well, which is included in the website above.
i wouldn't recommend Pascal's triangle for the first two. you'd end up having to count the terms as opposed to just plugging numbers into the binomial formula

using Pascal's triangle is probably easier for the last one, depending on your taste

4. You're right, I was just afradi he hadn't learned the binomial formula yet. Pascal's is just simpler in terms of the math needed, but will almost always take more time then using the binomial formula. They will both give you the same answer, but if you haven't learned the binomail formula, or don't want to learn it, Pascal's works.

5. Originally Posted by terr13
You're right, I was just afradi he hadn't learned the binomial formula yet. Pascal's is just simpler in terms of the math needed, but will almost always take more time then using the binomial formula. They will both give you the same answer, but if you haven't learned the binomail formula, or don't want to learn it, Pascal's works.
yes, i agree with you. i tend to prefer Pascal's method if i am expanding everything. but if they just ask me to pin-point one term, i find the binomial method to be easier.

since "Binomial expansion" was in the title, i assumed the poster would have some familiarity with it

6. Originally Posted by Jhevon
recall $(a + b)^n = \sum_{k = 0}^n {n \choose k} a^{n - k}b^k$

what does k have to be to get $x^6$?

use the same formula above, what does k have to be to get $x^3$

just expand based on the same formula again. you are writing out all the terms now

so for the first q does k=6 or 0?

thanks for all this help guys :] i have learnt it just rather brisk and have been so busy with work havent been keeping up

7. Originally Posted by coyoteflare
so for the first q does k=6 or 0?
you want the x-term to have a power of 6

look at the formula, and look at your problem. is the x-term represented by a or b? let this guide your choice for the answer

8. ## argh im lost

im lost... i got 6x^12

everything comes out with a power of 12?

9. ## so is it..

2^6 3x^6 ?

10. Originally Posted by coyoteflare
2^6 3x^6 ?
no.

we must have k = 6, since the b term represents the x-term in our problem

thus it is $1^{12 - 6} \cdot (3x)^6 = 729x^6$

(where did you get 2 from? we are doing $({\color{red}1} + 3x)^{12}$, right?)

EDIT: see post #13 for a slight correction to this

11. Originally Posted by Jhevon
no.

we must have k = 6, since the b term represents the x-term in our problem

thus it is $1^{12 - 6} \cdot (3x)^6 = 729x^6$

(where did you get 2 from? we are doing $({\color{red}1} + 3x)^{12}$, right?)
i thought it you had a (12/6) 1^12-6 x^6

12. Originally Posted by coyoteflare
i thought it you had a (12/6) 1^12-6 x^6
yes, you are right, i forgot that but it is ${12 \choose 6} = 924$ not $\frac {12}6 = 2$

remember, ${12 \choose 6} = _{12}C_6 = \frac {12!}{6! (12 - 6)!}$

so multiply my answer above by 924 (we want the coefficient, so drop the x^6 in your answer)

13. ## sorry for the double post.... had two win dows open and got confused..

i..im stressing alot over these!

so woudl the answer to the first one just be 1^12-6

baring in mind theres no calculators

14. Originally Posted by coyoteflare
i..im stressing alot over these!

so woudl the answer to the first one just be 1^12-6

baring in mind theres no calculators
i just did the first one

the coefficient of $x^6$ is ${12 \choose 6}1^{12 - 6} 3^6 = \frac {12!}{6!(12 - 6)!} \cdot 3^6 = \frac {3^612!}{2 \cdot 6!}$

15. Originally Posted by Jhevon
i just did the first one

the coefficient of $x^6$ is ${12 \choose 6}1^{12 - 6} 3^6 = \frac {12!}{6!(12 - 6)!} \cdot 3^6 = \frac {3^612!}{2 \cdot 6!}$