# Solve Binomial Theorem Homework: Find Coefficients of Degree 17 & x7

• Schaus
In summary, the coefficient of the term of degree 17 in the expansion of (x^2-x)^13 is -715. For (2x+3)^10, the coefficient of the term containing x^7 is 414720.
Schaus

## Homework Statement

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. Homework Equations

tk+1=nCkan-kbk

## 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!

Schaus said:
What am I being asked for with "the term of degree 17"?
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}##.

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

Schaus and Logical Dog
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.

Schaus said:
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.

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.

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

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.

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

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}.##

Schaus said:
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.
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.

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?

You're still making the same mistake as before.

*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.

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.

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

Looks good!

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

## 1. What is the Binomial Theorem?

The Binomial Theorem is a mathematical formula that allows us to expand binomial expressions, which are expressions with two terms, raised to a certain power. It is written as (a + b)^n, where a and b are the terms and n is the power.

## 2. How do you find the coefficients of a binomial expression?

To find the coefficients of a binomial expression, we use the Binomial Theorem formula, (a + b)^n, and plug in the values of the terms and power. Then, we use the combination formula, nCr, to find the coefficients. For example, to find the coefficients of (3x + 2)^5, we would use the formula (3x)^5 + (2)^5 = 243x^5 + 32, and then use the combination formula to get the coefficients: 1, 5, 10, 10, 5, 1.

## 3. What is the degree of a binomial expression?

The degree of a binomial expression is the highest power of the variable in the expression. For example, the degree of (3x + 2)^5 is 5, since the variable x is raised to the power of 5.

## 4. How do you find the coefficients of a specific degree in a binomial expression?

To find the coefficient of a specific degree in a binomial expression, we use the combination formula, nCr, where n is the power of the expression and r is the degree we are looking for. For example, to find the coefficient of x^3 in (3x + 2)^5, we would use the combination formula 5C3, which is equal to 10. Therefore, the coefficient of x^3 is 10.

## 5. Can the Binomial Theorem be used for expressions with more than two terms?

No, the Binomial Theorem can only be used for expressions with two terms. For expressions with more than two terms, we use the Multinomial Theorem, which is an extension of the Binomial Theorem for multiple terms.

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