Trial Q - Binomials (2 Viewers)

[krypticlemonjuice]

Can someone please show me how to do this q, the answer is D

 

[Drongoski]

Wow. Not straightforward. For only 1 mark?

Will post solution later, if I have the time. What puts me off is the amount of LaTeX input required.

 

[krypticlemonjuice]

Wow. Not straightforward. For only 1 mark?

Yep, would you be able to show how to do it please, if you don't mind?

 

[Drongoski]

Only term in x^3y^5 comes from, where i=1:

 

[howcanibesmarter]

So we can rearrange what's in the bracket to make it simpler: (1+y(x+y))^n

Since the question asks for the coeff for x^3y^5,

Expanding it using binomial theorem you will get something like
nC3 (y(x+y))^3 + nC4 (y^4(x+y)^4)

But you realise the only term that works is nC4 (y^4(x+y)^4 since the power adds up to 8.

Now since there is already a y^4 on the outside, we need a x^3y on the inside for nC4 (y^4(x+y)^4.

This gives you nC4 y^4 (4C1 x^3y). Multiplying it together gives you nC4*4C1 x^3y^5

Therefore the coefficient is nC4*4 (since 4c1 is just 4)

Hence, answer is D

Let me know if this doesn't make sense Do you need an explanation for q10 too? That one is pretty difficult if you've never come across the question before

 

[krypticlemonjuice]

So we can rearrange what's in the bracket to make it simpler: (1+y(x+y))^n

Since the question asks for the coeff for x^3y^5,

Expanding it using binomial theorem you will get something like
nC3 (y(x+y))^3 + nC4 (y^4(x+y)^4)

But you realise the only term that works is nC4 (y^4(x+y)^4 since the power adds up to 8.

Now since there is already a y^4 on the outside, we need a x^3y on the inside for nC4 (y^4(x+y)^4.

This gives you nC4 y^4 (4C1 x^3y). Multiplying it together gives you nC4*4C1 x^3y^5

Therefore the coefficient is nC4*4 (since 4c1 is just 4)

Hence, answer is D

Let me know if this doesn't make sense Do you need an explanation for q10 too? That one is pretty difficult if you've never come across the question before

Thank you so much for the explanation. I was confused how you expanded it though in the third line to get that. Can you explain how you did this please, as I didn't understand the following lines as much because of this

 

[krypticlemonjuice]

So we can rearrange what's in the bracket to make it simpler: (1+y(x+y))^n

Since the question asks for the coeff for x^3y^5,

Expanding it using binomial theorem you will get something like
nC3 (y(x+y))^3 + nC4 (y^4(x+y)^4)

But you realise the only term that works is nC4 (y^4(x+y)^4 since the power adds up to 8.

Now since there is already a y^4 on the outside, we need a x^3y on the inside for nC4 (y^4(x+y)^4.

This gives you nC4 y^4 (4C1 x^3y). Multiplying it together gives you nC4*4C1 x^3y^5

Therefore the coefficient is nC4*4 (since 4c1 is just 4)

Hence, answer is D

Let me know if this doesn't make sense Do you need an explanation for q10 too? That one is pretty difficult if you've never come across the question before

ALso yeah an explanation on 10 would be great!

 

[howcanibesmarter]

Thank you so much for the explanation. I was confused how you expanded it though in the third line to get that. Can you explain how you did this please, as I didn't understand the following lines as much because of this

So it's pretty much just binomial expansion. Do you know how to do it?

 

[krypticlemonjuice]

So it's pretty much just binomial expansion. Do you know how to do it?
View attachment 38825

Yeah I got that, but how did you get the nc3 and nc4 part ?? and why was it added

 

[howcanibesmarter]

ALso yeah an explanation on 10 would be great!

 

[howcanibesmarter]

Yeah I got that, but how did you get the nc3 and nc4 part ?? and why was it added

It's just part of the binomial theorem, the nc3 and nc4 is always there for any expansion, same as nc1, nc2, etc. It's another way to write pascal's triangle, or (n 3) ( except it's meant to vertical that's why i did nc3 instead)

 

[krypticlemonjuice]

@howcanibesmarter Thank you so much for your help!!