# Binomial Theorem

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1. Feb 11, 2017

### Schaus

1. The problem statement, all variables and given/known data
1. Given the binomial (x2-x)13determine the coefficient of the term of degree 17.
2. Given the binomial (2x+3)10 determine the coefficient of the term containing x7.

2. Relevant equations

tk+1=nCkan-kbk

3. The attempt at a solution
#1 - What am I being asked for with "the term of degree 17"? I tried using 17 as my K value but that doesn't work.

#2 - tk+1=nCkan-kbk
a=2x b=3 n=10 k=7
tk+1=10C7(2x)10-7(3)7
= (120)(8x3)(2187)
=2099520x3
Not even close to the answer I need. Any help would be greatly appreciated!

2. Feb 11, 2017

### andrewkirk

When you expand $(x^2-x)^{13}$ to standard polynomial form $\sum_{k=0}^{26}a_kx^k$, the term of degree 17 is $a_{17}x^{17}$. They are asking you what the value of $a_{17}$ is.
Hint: $(x^2-x)^{13}=x^{13}(x-1)^{13}$ so the answer will be the same as the coefficient of $x^{4}$ in the polynomial expansion of $(x-1)^{13}$.

3. Feb 11, 2017

### Staff: Mentor

#1: Write $x^2-x=x(x-1)$ first.
#2: Could it be, you exchanged $2$ and $3$.

4. Feb 11, 2017

### Schaus

tk+1=13C4(x2)13-4(-1)4. This is what I gather from what you guys said. If I use (-1)4 it turns my answer to a positive instead of negative. I tried switching a and b for the second question. I think there must be a typo in the answer sheet or learning guide.

5. Feb 12, 2017

### Ray Vickson

No, that is not a result of what people said. The learning guide is correct.

What is the coefficient of $x^4$ in the expansion of $(x-1)^{13}?$ There is no $x$ at all in the coefficient, because the coefficient of $x^4$ in the term $c_4 x^4$ is just $c_4$ itself. However, that is not the only error you wrote; another is more damaging.

6. Feb 12, 2017

### Schaus

Well whatever it is that I'm doing wrong, I'm unable to find it.

7. Feb 12, 2017

### Schaus

tk+1=10C3(2x)10-3(3)3. This might be what you were talking about for #2? When I did it like this I got the needed answer but I'm still at a loss for #1.

8. Feb 12, 2017

### Ray Vickson

The term containing $x^k$ in the expansion of $(x+a)^n$ is $C(n,k) a^{n-k} x^k$, so the coefficient is $C(n,k) a^{n-k}.$

9. Feb 12, 2017

### vela

Staff Emeritus
You seem to be mixing up two approaches.

Approach one: If $a=x^2$ and $b=x$, find the binomial expansion of $(a-b)^{13}$ and pick out the term that corresponds to $x^{17}$ when you substitute back in for $a$ and $b$.

Approach two: Pull out the common factor of $x$ first to get $(x^2-x)^{13} = [x(x-1)]^{13} = x^{13}(x-1)^{13}$. Use the binomial theorem to expand $(x-1)^{13}$ and then pick out the term that when multiplied by the other factor, $x^{13}$, results in the $x^{17}$ term.

10. Feb 12, 2017

### Schaus

tk+1=13C4(x2)13-4(-1)4
t5=13C4(x2)9(-1)4
t5=(715)(x18)(-1)4
=715x18
This is what I've come up with. Am I supposed to factor out the negative first before putting it all together?

11. Feb 12, 2017

### vela

Staff Emeritus
You're still making the same mistake as before.

12. Feb 12, 2017

### Schaus

*sigh* I figured as much. Something just isn't clicking for me on this and I just don't understand what I'm doing wrong.

13. Feb 12, 2017

### vela

Staff Emeritus
For one thing, you're asked to find the coefficient of $x^{17}$, but you have $x^{18}$. Note that your expression will always produce an even power of $x$. That's a clue as to where the mistake might lie. It's really a matter of doing things step by step and paying close attention to the details. When you write an expression down, go through it piece by piece and make sure it's correct.

14. Feb 12, 2017

### Schaus

I found that t10=(13C9)(x2)13-9)(-x)9
t10=(715)(x2)4)(-x)9
=(715)(x8)(-x)9
=-715x17?

15. Feb 12, 2017

### vela

Staff Emeritus
Looks good!

16. Feb 12, 2017

### Schaus

Awesome thanks for the help. Now I just hope I can reproduce this result with another question!