1. ## Binomial Expansion help.

Hey,

The question is:

For the binomial expansion of ascending powers of x of (1+1/4x)^n, when n is an integer and n is bigger than 2.

Find and simplify the first three terms.

I got it to be 1+1/4x+(1/4x)^2

But the answers are something different.

Any idea what Im doing wrong?

Thanks

2. Pascals Triangle should help....

For n=2,

$\displaystyle 1+2/4x+(1/16)x^2$

Because it is 2 * 1/4

(a+b)^2 = a^2 + 2ab + b^2

BUT, n is bigger than 2, so it should be...

n = 3
$\displaystyle (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Pascal's Triangle is the one for probability, but those numbers also work for binomial expansions.

Good luck.

3. When its n is bigger than 2 I dont think you can assign n as 3 it has to be n. If that makes sense?

4. $\displaystyle \left( {a + b} \right)^n = \sum\limits_{k = 0}^n {\binom{n}{k}a^k b^{n - k} }$

5. Oh, no, I meant it as an example.

So, looking at pascal's triangle for n = 4 (the fourth row), what is the expansion?

I think that what is being said above me is waaay to complicated for you, but is saying that you do the sum of the Combinations nCr for the numbers you are given, and that is Pascal's triangle.

Read a short article on it, the basics at least, to understand the expansion with ease.

6. I do understand what your saying and I have learnt about pascals triangle and using the nCp key but what I was hoping for was if someone were to attempt the question and show me where I have gone wrong.

7. $\displaystyle ^nC_k=\binom{n}{k}$.
$\displaystyle (x+y)^6=\binom{6}{0}x^{6}y^{0}+\binom{6}{1}x^{5}y^ {1}+\binom{6}{2}x^{4}y^{2}+\binom{6}{3}x^{3}y^{3}+ \binom{6}{4}x^{2}y^{4}+\binom{6}{5}x^{1}y^{5}+\bin om{6}{6}x^{0}y^{6}$

8. I understand that Plato, but it doesn't really help.

9. Originally Posted by jamesg007
I understand that Plato, but it doesn't really help.
Then what in the world are you asking?

10. Try reading my first post.

11. Originally Posted by jamesg007
Why would you want to use Pascal's triangle to find the coefficients for? It is grossly inefficient to use it for high powers of n.

You can also use the formula

$\displaystyle (1+x)^n = 1+nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3$$\displaystyle + \frac{n(n-1)(n-2)(n-3)}{4!}x^4 + ... + \frac{n!}{n!}x^n$

The coefficients of k are $\displaystyle {n\choose k} = \frac{n!}{(n-k)!k!}$

12. Thanks

13. Originally Posted by jamesg007
Im not the one advocating the use of Pascals triangle.
My apologies

$\displaystyle \left(1+ \frac{1}{4x}\right)^n$

You can use the formula in my last post but where $\displaystyle x = \frac{1}{4x}$ in this case

### expand (2 1/4x)^9 in ascending powers

Click on a term to search for related topics.