Help Binomial Question (2 Viewers)

[notme123]

View attachment 32074

Apologies for my working out, I just quickly rushed solving the problem.

For some faded parts:
In line 2: - 10!/k!*(10-k)!, in this line i also cancelled out the negatives

On that note, I also struggle with this topic a bit so if anyone in the forum could answer this;

How do you know which term is a and b in the binomial expansion? Sometimes the formula works if the a term is the coefficient of x and vice versa. I'm confused by that so if someone could clarify that, it would be nice.

oh i see my denominators are switched for some reason
thank you so much

 

[notme123]

thank you so much

View attachment 32074

Apologies for my working out, I just quickly rushed solving the problem.

For some faded parts:
In line 2: - 10!/k!*(10-k)!, in this line i also cancelled out the negatives

On that note, I also struggle with this topic a bit so if anyone in the forum could answer this;

How do you know which term is a and b in the binomial expansion? Sometimes the formula works if the a term is the coefficient of x and vice versa. I'm confused by that so if someone could clarify that, it would be nice.

with the first line, whats the reason you can take out a negative?
oh dont worry its because you want highest magnitude

 

[CM_Tutor]

how else can this question be asked? it seems the method would always be the same

What I meant was that some people like to use a memorised formula for the ratio of terms in the binomial expansion of , which is

​

but this can be prevented by writing a question as something like:

Consider the binomial expansion

​

Find an expression for and hence find the prime factorisation of the greatest term in the expansion when .

 

[CM_Tutor]

When looking at the HSC past questions and Baulkham Hill's 2020 year 11 ext yearly the types of ways the question can be asked are:
1. constant term
2. coefficient of specific index
3. coefficient of x^3 in (1 + x + x^2)^5
4. I don't even know how to summarise the next question.
View attachment 32064

Your "type 4" is an example of seeking a relationship between binomial coefficients.

 

[CM_Tutor]

how exactly? i know the binomial expansion if thats what youre referring to

Looking at we have the same expression multiplied 5 five times and each term must include exactly one term from each set of parentheses. To get a term in , I can select:

• 3 of the term "" and two of the term 1 in ways
• 1 of the term "" and one of the term "" plus three of the term 1 in ways

So, the coefficient of is 10 + 20 = 30.

 

[notme123]

Looking at we have the same expression multiplied 5 five times and each term must include exactly one term from each set of parentheses. To get a term in , I can select:

• 3 of the term "" and two of the term 1 in ways
• 1 of the term "" and one of the term "" plus three of the term 1 in ways

So, the coefficient of is 10 + 20 = 30.

thats so clever

 

[ExtremelyBoredUser]

Ya, those type of questions are my favourites for binomials, really forces you to think about the process rather than just following a formula.

 

[notme123]

Ya, those type of questions are my favourites for binomials, really forces you to think about the process rather than just following a formula.

Here's another one I solved. Find the x^4 coefficient of

 

[ExtremelyBoredUser]

Here's another one I solved. Find the x^4 coefficient of

Nice, I'll check it out after I finish my homework. My first idea just by looking at the question is to decompose it into a binomial expression and then input it into the binomial formula and solve from there.

 

[notme123]

Nice, I'll check it out after I finish my homework. My first idea just by looking at the question is to decompose it into a binomial expression and then input it into the binomial formula and solve from there.

you can use the previous method CMtutor did. But theres a slight difference.

 

[ExtremelyBoredUser]

you can use the previous method CMtutor did. But theres a slight difference.

Sorry for my late reply, I've had some internet issues. I've done the problem and I'm probably wrong so is the coeff 83160? I think Im wrong because I didnt find anything tricky with the question or out of the blue with those type of questions, is there something I've overlooked?

 

[Undefined]

Looking at we have the same expression multiplied 5 five times and each term must include exactly one term from each set of parentheses. To get a term in , I can select:

• 3 of the term "" and two of the term 1 in ways
• 1 of the term "" and one of the term "" plus three of the term 1 in ways

So, the coefficient of is 10 + 20 = 30.

That way is very clever. Just for comparison, I will post the solutions to the Baulkham Hills paper I quoted it from. Took me 10 minutes with the wrong solution when I tried to do the question half an hour ago after warming myself up with easier binomial expansion questions

 

[ExtremelyBoredUser]

That way is very clever. Just for comparison, I will post the solutions to the Baulkham Hills paper I quoted it from. Took me 10 minutes with the wrong solution when I tried to do the question half an hour ago after warming myself up with easier binomial expansion questions
View attachment 32080

Indeed, this is the more conventional method of finding the coefficient and I'm sure most people are trained to do this way but I really liked CM_Tutor's way more since it feels more intuitive and true to the topic of combinatorics. This isn't a slow method by any means if you practice it, it should take 1-2 mins at most.

 

[CM_Tutor]

Here's another one I solved. Find the x^4 coefficient of

Sorry for my late reply, I've had some internet issues. I've done the problem and I'm probably wrong so is the coeff 83160? I think Im wrong because I didnt find anything tricky with the question or out of the blue with those type of questions, is there something I've overlooked?

I get the same answer, and it is confirmed by the online trinomial expansion calculator at https://www.symbolab.com/solver/expand-calculator/expand \left(3+4x+5x^{3}\right)^{6}

Spoiler

• There are ways to get a term of the form , contributing a term of
• There are ways to get a term of the form , contributing a term of

I see no other way to get a term in , making the term

​

 

[ExtremelyBoredUser]

I get the same answer, and it is confirmed by the online trinomial expansion calculator at https://www.symbolab.com/solver/expand-calculator/expand \left(3+4x+5x^{3}\right)^{6}

Spoiler

• There are ways to get a term of the form , contributing a term of
• There are ways to get a term of the form , contributing a term of

I see no other way to get a term in , making the term

​

Oh neat there's a website for that, I'll use that in the future to double check my answers. Thanks for showing another way to think of the problems as well, it helps saves some time.