Binomial Theorems - Can Someone Please Explain What The Question Is Asking?

[knpoe03]

Hello!

As per the requests of past Helpers, I will be including an image below rather than a PDF. I am unsure of what exactly the worksheet is asking me to do in question 1c. Would someone mind explaining to me how I should approach this, or breaking down what exactly the question wants me to do?

The writing in red is numbers that I have chosen. I chose these numbers at random, just whatever sounded best to me.

Thanks!

NOTE: Before someone tells me that I need to post my work, (no work=no help) I honestly have no idea of how to go about this. I do not have any work to provide you with because I don't even know how to go about attempting this question.

 

[Romsek]

Binomial Theorem

[MATH](a x^p - b y^q)^{25} = \sum \limits_{k=0}^{25}\dbinom{25}{k} (a x^p)^k (-b y^q)^{25-k}[/MATH]
For the 7th term let [MATH]k=6[/MATH]

 

[JeffM]

[MATH](a x^p - b y^q)^{25} = \sum \limits_{k=0}^{25} \dbinom{25}{k} * (a x^p)^k (-b y^q)^{25-k}[/MATH]

 

[knpoe03]

[MATH](a x^p - b y^q)^{25} = \sum \limits_{k=0}^{25} \dbinom{25}{k} * (a x^p)^k (-b y^q)^{25-k}[/MATH]

Binomial Theorem

[MATH](a x^p - b y^q)^{25} = \sum \limits_{k=0}^{25}\dbinom{25}{k} (a x^p)^k (-b y^q)^{25-k}[/MATH]
For the 7th term let [MATH]k=6[/MATH]

Oh, that makes a lot more sense! I'm going to go try that out and see how it works. I can post my work in a bit. Thank you!!

 

[knpoe03]

@JeffM @Romsek Gosh, I'm stuck again. I don't know to do once I've determined this:

 

[Steven G]

Just compute the right hand side when k=7. You should not have a summation symbol and between the parenthesis, ( ), for the combination you should not have a division sign.

Note that the right hand side says to add up many terms (26 in your case) and you are asked to find just one of those terms so there is no need to have the summation sign any more. Now since the right hand side is just one of the 26 terms it can not equal the left hand side any more.

[MATH] \binom{25}{k} * (a x^p)^k (-b y^q)^{25-k}[/MATH] Just replace k with 6, a with 4 and b with 6 simplify and you'll be done

 

[JeffM]

I can't say that I would give an A+ to the way the question is worded, but you can get there if you work at it a bit.

First, there is no problem to work on with

[MATH](4x^8 - 6y^{12})[/MATH] because (a) there is no seventh term, and

(b) IT IS FALSE that [MATH](4x^8 - 6y^{12}) = \sum_{k=0}^{25} \dbinom{25}{k} * ((ax)^p)^k * ((by)^q)^{25-k}.[/MATH]
What is true is [MATH](4x^8 - 6y^{12})^{25} = \sum_{k=0}^{25} \dbinom{25}{k} * ((ax)^p)^k * ((by)^q)^{25-k}.[/MATH]
The expansion of that sum has 26 terms. Each term contains 3 factors. A numeral, x exponentiated, and y exponentiated.

Part c asks what the exponents are.

Part d asks what is the numeral

 

[knpoe03]

Just compute the right hand side when k=7. You should not have a summation symbol and between the parenthesis, ( ), for the combination you should not have a division sign.

Note that the right hand side says to add up many terms (26 in your case) and you are asked to find just one of those terms so there is no need to have the summation sign any more. Now since the right hand side is just one of the 26 terms it can not equal the left hand side any more.

[MATH] \binom{25}{k} * (a x^p)^k (-b y^q)^{25-k}[/MATH] Just replace k with 6, a with 4 and b with 6 simplify and you'll be done

I can't say that I would give an A+ to the way the question is worded, but you can get there if you work at it a bit.

First, there is no problem to work on with

[MATH](4x^8 - 6y^{12})[/MATH] because (a) there is no seventh term, and

(b) IT IS FALSE that [MATH](4x^8 - 6y^{12}) = \sum_{k=0}^{25} \dbinom{25}{k} * ((ax)^p)^k * ((by)^q)^{25-k}.[/MATH]
What is true is [MATH](4x^8 - 6y^{12})^{25} = \sum_{k=0}^{25} \dbinom{25}{k} * ((ax)^p)^k * ((by)^q)^{25-k}.[/MATH]
The expansion of that sum has 26 terms. Each term contains 3 factors. A numeral, x exponentiated, and y exponentiated.

Part c asks what the exponents are.

Part d asks what is the numeral

Thank you all for your comments. So sorry for my confusion, this is our first assignment on this subject and I'm completely lost. I will include an updated photo of my work below.

So, I've fixed the mistakes in my work and set it up according to Mr. @Jomo and Mr. @JeffM . I have a few questions:
1) I've never seen a fraction thingy before without the fraction line. I assumed that it was some sort of glitch, but that it was a fraction. Now that I know it's not a fraction, but I don't know how else to describe it. It is at the beginning of the problem with the 25 over the k. What is this called?
2) For simplifying the equation, do I just complete the (4x^8)^6 and the (-6y^12)^25-6, then multiply both of these together?

Once again, I'm super sorry. I've never encountered anything like this and I am entirely lost.

 

[Dr.Peterson]

1) I've never seen a fraction thingy before without the fraction line. I assumed that it was some sort of glitch, but that it was a fraction. Now that I know it's not a fraction, but I don't know how else to describe it. It is at the beginning of the problem with the 25 over the k. What is this called?

This is a reason why, if you have no work to show, you should at least show what you have been taught, perhaps an example, so we can see what notation you have learned, and then use that in our answers. In this case, that would mean showing what the binomial theorem means, as taught to you.

The notation [MATH]{{n}\choose{r}}[/MATH] is called the binomial coefficient, and is the same thing as [MATH]_nC_r[/MATH], the number of combinations of n items taken r at a time. Is any of that familiar to you?

 

[JeffM]

Expanded out, what is [MATH](4x^8)^6[/MATH]? Use exponential notation for the numeral because some of these are going to be very big numbers..

Expanded out, what is [MATH](-6y^{12})^{25-6}[/MATH]? Again, use exponential notation for the numeral.

Now, as Dr. Peterson explained, [MATH]\dbinom{n}{r} \equiv [/MATH] _nC_r.

To do this assignment, you need the the formula for calculating the numeric value of these equivalent symbols. It should be in your text or notes.

 

[knpoe03]

This is a reason why, if you have no work to show, you should at least show what you have been taught, perhaps an example, so we can see what notation you have learned, and then use that in our answers. In this case, that would mean showing what the binomial theorem means, as taught to you.

The notation [MATH]{{n}\choose{r}}[/MATH] is called the binomial coefficient, and is the same thing as [MATH]_nC_r[/MATH], the number of combinations of n items taken r at a time. Is any of that familiar to you?

I have no notes. My teacher uses a reverse system, where we have to do the work first and figure it out for ourselves, THEN she will give us a lesson. It doesn't help that my school has been forced to go virtual, either. So, I'm pretty much struggling through this alone.

I've never heard of a binomial coefficient or of the nCr thing. I will have to do research on that.

 

[knpoe03]

Expanded out, what is [MATH](4x^8)^6[/MATH]? Use exponential notation for the numeral because some of these are going to be very big numbers..

Expanded out, what is [MATH](-6y^{12})^{25-6}[/MATH]? Again, use exponential notation for the numeral.

Now, as Dr. Peterson explained, [MATH]\dbinom{n}{r} \equiv [/MATH] _nC_r.

To do this assignment, you need the the formula for calculating the numeric value of these equivalent symbols. It should be in your text or notes.

I do not have notes. I'll copy this from my previous comment: "I have no notes. My teacher uses a reverse system, where we have to do the work first and figure it out for ourselves, THEN she will give us a lesson. It doesn't help that my school has been forced to go virtual, either. So, I'm pretty much struggling through this alone."

I think I understand some of what you are saying. I'm going to go attempt to do what you said, and I'll get back to you with my updated work.

 

[JeffM]

I have no notes. My teacher uses a reverse system, where we have to do the work first and figure it out for ourselves, THEN she will give us a lesson. It doesn't help that my school has been forced to go virtual, either. So, I'm pretty much struggling through this alone.

I've never heard of a binomial coefficient or of the nCr thing. I will have to do research on that.

It's a simple formula.

Binomial coefficient - Wikipedia

en.wikipedia.org

The definition of the factorial is as follows

[MATH]n! = 1 \text { if } n = 0\\ n! = n * (n - 1)! \text { if } n > 0.[/MATH]What this means is that

[MATH]n \ge 2 \implies n! = \prod_{k=2}^n k.[/MATH]
So for example [MATH]5! = \prod_{k=2}^5 k = 2 * 3 * 4 * 5 = 120 = 5 * 4 * 3 * 2 * 1.[/MATH]

 

[knpoe03]

@JeffM Am I close? If so, would m be 48, n be 228, and coefficient be 4? Or, do I need to multiply the second and third sections (within the parenthesis) together?

 

[JeffM]

Let's take this in steps. The notation is making you slip up.

Yes, m = 48, but it is n = 228 rather than y.

Next, I'd not approximate [MATH](-6)^{19}[/MATH] alone. I do not like approximations until I know what accuracy is required. Here we can give an exact answer. Your teacher may have a different view. And for ease in the final step, I'd probably break those coefficients down into powers of prime factors..

Finally, [MATH]\dbinom{25}{6} * 4^6 * (-6)^{19} = - \dbinom{25}{6} * 2^{31} * 3^{19} = WHAT?[/MATH]

 

[Steven G]

Note that maybe (-6)¹⁹= -6¹⁹, but if the power is even like in (-6)²⁰ it does not equal -6²⁰. That is you just can't simply put the negative sign in front and then raise the number to the given power. For the record, (-6)²⁰ = 6²⁰. Why? Because if you multiply 20 (an even number) negative numbers you will get a positive answers. If you multiply 19 negative numbers you will get a negative answer.

 

[knpoe03]

Let's take this in steps. The notation is making you slip up.

Yes, m = 48, but it is n = 228 rather than y.

Next, I'd not approximate [MATH](-6)^{19}[/MATH] alone. I do not like approximations until I know what accuracy is required. Here we can give an exact answer. Your teacher may have a different view. And for ease in the final step, I'd probably break those coefficients down into powers of prime factors..

Finally, [MATH]\dbinom{25}{6} * 4^6 * (-6)^{19} = - \dbinom{25}{6} * 2^{31} * 3^{19} = WHAT?[/MATH]

Do I leave the [MATH]\dbinom{25}{6}[/MATH] out? If so, would it be [MATH]\dbinom{25}{6} * 6^{589}[/MATH]? (I multiplied the 2 and 3, then multiplied their exponents. I feel like this is probably wrong.)

EDIT: Is it actually equal to (4x^8−6y^12)^25?

DOUBLE EDIT: But where do the x and y go?

 

[Steven G]

No you do not multiply the 2 and 3 together because you do NOT have a 2 and 3 to multiply. You have a 2¹³ and a 3¹⁹. Before you multiply you need to know what you are multiplying! Again, you are to multiply 2¹³ and 3¹⁹.
This is my style, but I would not simplify as JeffM did, I would just multiply the three factors you had.

You need to compute [math]\binom {25}{6}[/math], 4⁶ and (-6)¹⁹, multiply does three values together and say done.[

 

[knpoe03]

No you do not multiply the 2 and 3 together because you do NOT have a 2 and 3 to multiply. You have a 2¹³ and a 3¹⁹. Before you multiply you need to know what you are multiplying! Again, you are to multiply 2¹³ and 3¹⁹.
This is my style, but I would not simplify as JeffM did, I would just multiply the three factors you had.

You need to compute [math]\binom {25}{6}[/math], 4⁶ and (-6)¹⁹, multiply does three values together and say done.[

@JeffM @Jomo I have a few questions:
1) How do I compute the (25 over 6)?
2) When I multiply these values, should I keep the exponents?
3) After multiplying these values, will it provide me with the m, n, and coefficient? Or, did we find the m and n earlier? (For reference, I said that m was 48 and n was 228. The worksheet is at the top of the page so that you can see what it says I need to find.)

 

[knpoe03]

@JeffM @Jomo After multiplying the three terms together (used my calculator) would the final solution be -725401600 * 6^19? If so, would the 6 be the coefficient?