# Binomial Expansions.

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1. Feb 9, 2016

### Meezus

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
After using this formula I got 1-3x + 6x2 for (1-x)-3 and 1+x - 1/8 ∙ 4x2 for (1+2x)1/2 for the second part of the question I'd assume that you're supposed to multiply the equations? I think this done by timing the whole of second bracket by first number (1) then second number (-3x) and then third number (6x2.from the second bracket.

2. Feb 9, 2016

### blue_leaf77

Check again the sign of the second term.

3. Feb 9, 2016

### Meezus

I don't see where I'm going wrong here? (-3)(-4)(x)2(1/2) = 6x2

4. Feb 9, 2016

### blue_leaf77

The second term, not the third one.

5. Feb 9, 2016

### PeroK

No one says "timing". The word is "multiplying". But, yes, that's just the distributive law.

6. Feb 9, 2016

### Meezus

Sorry, I'm not seeing the mistake? -3(x) = -3x correct?

7. Feb 9, 2016

### blue_leaf77

The formula for the binomial expansion uses a plus sign in front of $x$ to define the equation: $(1+x)^n$. While in the question, the sign of $x$ is minus.

8. Feb 9, 2016

### Meezus

Oh I was just looking at it as +(-3) which is same as just -3 so I can't use this formula?

9. Feb 9, 2016

### Meezus

My mistake, would you mind checking to see if I did the first part correct?

10. Feb 9, 2016

### blue_leaf77

You can use that formula, but the sign of each $x$ appearing in the RHS (the infinite sum) must be adjusted accordingly. I will rewrite the formula using different variable notation for the sake of clarity
$$(1+u)^{-3} = 1-3u+6u^2-10u^3+\ldots$$
what if you change $u\rightarrow -x$ as you have in the problem?

11. Feb 9, 2016

### Meezus

ooohh it all becomes negative? I see! So its 1-3x - 6x2?

12. Feb 9, 2016

### blue_leaf77

Ok rather than an arrow sign, I will use an equality sign. Make a change $u=-x$ in that formula. Does it help this time?

13. Feb 9, 2016

### PeroK

It's not the binomial theorem that is the problem, it's your basic algebra, especially handling negative signs. I suggest you do a bit of revision of this.

14. Feb 9, 2016

### Meezus

1-3-x + 6-x2 - 10-x3

15. Feb 9, 2016

### blue_leaf77

I know what you did there is simply replacing $u$ with $-x$ without giving heed to the rules of writing the accepted mathematical expression. What does -3-x in that expression mean?

16. Feb 9, 2016

### Meezus

I'm not sure how it should be expressed. -3-x = -(x+3). Should it be 1 - (3+x)? (IGNORE)

Its (-3)(-x) so it should be 3x because its -3 * -x. sorry

17. Feb 9, 2016

### blue_leaf77

Yes, it should have been $+3x$.