The Student Room Group

C2 binomial expansion

https://b940eb267c31463d4d23ccfd3105c3b78e5616c6.googledrive.com/host/0B1ZiqBksUHNYUnQzSHRkZ3JoNGs/for-Edexcel/Solomon%20B%20QP%20-%20C2%20Edexcel.pdf

help with 3a with the 3rd term in the expansion please.

so far i have 1.(x4)1.\left(\dfrac{x}{4}\right)

not sure about the power which goes on the x/4 and how to express nC2^n\mathrm{C}_2 as a coefficient
Reply 1
Original post by thefatone
https://b940eb267c31463d4d23ccfd3105c3b78e5616c6.googledrive.com/host/0B1ZiqBksUHNYUnQzSHRkZ3JoNGs/for-Edexcel/Solomon%20B%20QP%20-%20C2%20Edexcel.pdf

help with 3a with the 3rd term in the expansion please.

so far i have 1.(x4)1.\left(\dfrac{x}{4}\right)

not sure about the power which goes on the x/4 and how to express nC2^n\mathrm{C}_2 as a coefficient

I recommend you use the formula given in the Edexcel formula book for:

(1+x)n\displaystyle (1+x)^n
Reply 2
Original post by notnek
I recommend you use the formula given in the Edexcel formula book for:

(1+x)n\displaystyle (1+x)^n


i see... are you supposed to learn about that or is it just given and you know how to use it?
Reply 3
Original post by thefatone
i see... are you supposed to learn about that or is it just given and you know how to use it?


Both?
Reply 4
Original post by Zacken
Both?


... so how would you get the 3rd term of the expansion of (1+x)n\left(1+x\right)^n

i'm not sure about this i haven't been taught this :/
Reply 5
Original post by thefatone
... so how would you get the 3rd term of the expansion of (1+x)n\left(1+x\right)^n

i'm not sure about this i haven't been taught this :/


It's given in the booklet, it's n(n1)(n2)3!x3\frac{n(n-1)(n-2)}{3!} x^3
Reply 6
Original post by Zacken
It's given in the booklet, it's n(n1)(n2)3!x3\frac{n(n-1)(n-2)}{3!} x^3


would i ever need to "show" that ?
Reply 7
Original post by thefatone
would i ever need to "show" that ?


No.But it's not hard: nC3=n!(n3)!3!=n(n1)(n2)(n3)!(n3)!3!=n(n1)(n2)3!\displaystyle ^n\mathrm{C}_3 = \frac{n!}{(n-3)! 3!} = \frac{n(n-1)(n-2)(n-3)!}{(n-3)! 3!} = \frac{n(n-1)(n-2)}{3!}
Reply 8
Original post by thefatone
https://b940eb267c31463d4d23ccfd3105c3b78e5616c6.googledrive.com/host/0B1ZiqBksUHNYUnQzSHRkZ3JoNGs/for-Edexcel/Solomon%20B%20QP%20-%20C2%20Edexcel.pdf

help with 3a with the 3rd term in the expansion please.

so far i have 1.(x4)1.\left(\dfrac{x}{4}\right)

not sure about the power which goes on the x/4 and how to express nC2^n\mathrm{C}_2 as a coefficient


if you did c4 this question would have made more sense, binomial questions to the power of N has never come up on edexcel papers. But as stated below you would need to do the long method of working it out
Original post by thefatone
https://b940eb267c31463d4d23ccfd3105c3b78e5616c6.googledrive.com/host/0B1ZiqBksUHNYUnQzSHRkZ3JoNGs/for-Edexcel/Solomon%20B%20QP%20-%20C2%20Edexcel.pdf

help with 3a with the 3rd term in the expansion please.

so far i have 1.(x4)1.\left(\dfrac{x}{4}\right)

not sure about the power which goes on the x/4 and how to express nC2^n\mathrm{C}_2 as a coefficient


finding and simplifying in part a should still be in terms of n, then for part be you can use that to answer
Reply 10
Original post by Zacken
No.But it's not hard: nC3=n!(n3)!3!=n(n1)(n2)(n3)!(n3)!3!=n(n1)(n2)3!\displaystyle ^n\mathrm{C}_3 = \frac{n!}{(n-3)! 3!} = \frac{n(n-1)(n-2)(n-3)!}{(n-3)! 3!} = \frac{n(n-1)(n-2)}{3!}


I see thanks a ton :smile:
wait when does n factorial expand to all those brackets????
Reply 11
Original post by thefatone
I see thanks a ton :smile:
wait when does n factorial expand to all those brackets????


What does the factorial represent?

Unparseable latex formula:

\displaystyle [br]\begin{equation*}n! = n\underbrace{(n-1)(n-2)(n-3)\cdots (2)(1)}_{(n-1)!}\end{equation*}



Or

Unparseable latex formula:

\displaystyle [br]\begin{equation*} n!=n(n-1)\underbrace{(n-2)\cdots (2)(1)}_{(n-2)!} = n(n-1)(n-2)! \end{equation*}



Or...

It's not hard if you just think for a bit.
(edited 7 years ago)
Original post by thefatone
I see thanks a ton :smile:
wait when does n factorial expand to all those brackets????


imagine 5! ... 5x4x3 etc... minusing 1 each time
Reply 13
Original post by Zacken
What does the factorial represent?

Unparseable latex formula:

\displaystyle [br]\begin{equation*}n! = n\underbrace{(n-1)(n-2)(n-3)\cdots (2)(1)}_{(n-1)!}\end{equation*}



Or

Unparseable latex formula:

\displaystyle [br]\begin{equation*} n!=n(n-1)\underbrace{(n-2)\cdots (2)(1)}_{(n-2)!} = n(n-1)(n-2)! \end{equation*}



Or...

It's not hard if you just think for a bit.


thanks

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