A-Level Maths: D1-17 [Binomial Expansion: Find the coefficient of x^10 in (1-2x)^5*(2+x)^7]

[Durag_sensei]

Thank you so much I’ve just started my a levels and online school has been extremely hard but your videos have made everything so much easier and better to understand

[kiangregory]

Im stressing over AS and youre a lifesaver thank you sm! You explain things way better than my actual teachers

[musaa]

My good sir, you deserve recognition for effective communication. Not a single moment of confusion here. Give a Ted Talk on maths sometime.

[aminamiah]

Hi Sir, I found this video quite hard to understand when it came to finding the coefficients, why did you start with 1 to the power of 2 and -2x to the power of 3?

[christyjane]

Hi sir, these videos are great and I hope the next assessment I will ace them but I have a question do have any useful links for exam-style

[ebanybolokoli]

Does this method work for when one of the brackets doesn't have any indicies? e.g find the coefficient of x^3 in the binomial expansion (2+x)(3-2x)^7

[hypedmaniac]

Tried this another way before watching the video. It wound up being wrong despite making sense in my head - what’s the error here?
(Going to use * instead of x here for multiplication - too many x’s being thrown around otherwise)

By law of indices, 7+5 = 12, so n=12, r=10.
So 12C10 * ((1^(12-10))*(2^(12-10)) * (((2x)^5)*(x^7)) = -8448x^10
So I found the coefficient to = -8448, which was wrong.

[abdullahhasan]

shouldnt the 2nd coefficient of x^10 that u have calculated be negative?? like the other two?? so instead of 1120, it should be -1120 and the final answer should be -3888???

[ronal]

in the second term shouldn't you factor 2 out of the equation to make it to the form of 2^7[1+x/2]^7

[Muslim]

can't there be like more possibilities of getting x^10 like ( x^2 * x^8 ) or (x^9 * x^1) thats the only thing which scratches my mind

[joshuacarpenter]

Do you recommend I take further maths for a degree in engineering?

[somedumbusername]

I’m in GCSE rn but I’m applying for a scholarship to a private school for math and need to know this for the test. 😭

[laibaimtinan]

Not all heroes wear capes, some wear hoodies .

[van]

Whilst not quite the same as the problem you have gone through I have a couple of questions about a similar sort of problem.

Looking at the Edexcel Formula Book if provides two definitions of the Binomial expansion (which (sadly) are near impossible to enter into this comment form!)

The first is the expansion of (a + b) ^n where n belongs to the set of N (Natural Nos).

The second definition (which you can obtain from the first) is the expansion of (1 + x) ^n where mod(x) < 1 and n belongs to the set of real no's R.

My first question in the derivation of the second form of the expansion (i.e. substituting 1 = a and x = b) into definition 1, there is nowhere I observe the condition of mod(x) < 1 or indeed n being a real no. materialising. Can you provide an insight into where these conditions emanate from?

My second question is associated with the practical implementation of the second definition. Given in the second definition states n belongs to the set of real numbers then this therefore supports both the use of fractional and negative indices.

So, given the problem: State the first four terms of A = (2 + x)^3 divided by SQRT(4 + x) then

A = ( (2 + x)^3 ) * ( (4 + x)^-0.5 )



1. assuming you can expand both terms separately (using the second definition) - I assume this is valid (please keep me honest if I am wrong)
2. then distil out the first four terms from the product of both resulting expansions
3. You obtain the answer.

However, my question is as follows: What is the resulting valid range?

Given the second definition states mod(x) < 1 for the binomial expansion then

1. For the first expansion (2 + x)^3 has a range of mod(x/2) < 1 i.e. mod(x) < 2
2. Whereas the second expansion SQRT(4 + x) has a range mod(x/4) < 1 i.e. mod(x) < 4

Therefore which is the correct Range to state and why so?

Apologies for length of this post - mould be much easier if I could have sent you a screenshot or similar. Is there any means of being able to send something like that to you in future?

Many thanks for any help you can offer

VBR

[shaziekhan]

How are there 0 likes in this video !?!?!?

[md.s.a.e]

So why didn't you completed the whole video.