# Thread: Binomial Expansion

1. ## Binomial Expansion

We started doing binomial expansions today and I really do not get it at all. I get that in the form:

(a+b)^n the exponents on a decrease and on b they increase and that there are n + 1 terms and you use pascal's triangle to get the terms but I don't quite get how to simplify what I come up with.

E.g. (3x+4a)^5
= 3x^5 + 5(3x^4+4a^1) + 10(3x^3+4a^2) + 10(3x^2+4a^3) + 5(3x+4a^4) + (4a^5)

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

This is as far as I get but I don't get how to simplify it. Any help would be super awesome. I don't even know if that is correct.

Thanks.

2. Your expansion was not quite right, this is how is should be

$\displaystyle (3x+4a)^5= (3x)^5 + 5(3x)^4\times (4a)^1 + 10(3x)^3\times(4a)^2 + 10(3x)^2\times(4a)^3 +$ $\displaystyle 5(3x\times(4a)^4) + (4a)^5$

Here's some help with the first 2 terms.

$\displaystyle (3x+4a)^5= 3^5x^5 + 5(3^4x^4\times 4a) +\dots$

$\displaystyle (3x+4a)^5= 243x^5 + 5(81x^4\times 4a) +\dots$

$\displaystyle (3x+4a)^5= 243x^5 + 405x^4\times 4a +\dots$

$\displaystyle (3x+4a)^5= 243x^5 + 1620x^4a +\dots$

3. Thanks,

when I expand it I get this:

$\displaystyle 243x^5 + 1620x^4a + 4320x^3a^2 + 5760x^2a^3 * 3840xa^4 + 1024a^5$

Does that look right?

4. $\displaystyle 243x^5 + 1620x^4a + 4320x^3a^2 + 5760x^2a^3 + 3840xa^4 + 1024a^5$

is correct, be careful with your signs.