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C2 Binomial Expansion

Hi there,

I've never seen this type of question before and I am very confused...

So I know that nC0 = 1
and that nC1 = n

So I have 1 + n(x/4) as the first two terms

I have no idea where go from there - This topic was only recently taught and I'm not very good at it, so if anyone could provide a detailed explanation on how to tackle this question I would be most grateful.

Thanks
Reply 1
Avatar for Zacken
Zacken
22
Original post by CrazyFool229
Hi there,

I've never seen this type of question before and I am very confused...

So I know that nC0 = 1
and that nC1 = n

So I have 1 + n(x/4) as the first two terms

I have no idea where go from there - This topic was only recently taught and I'm not very good at it, so if anyone could provide a detailed explanation on how to tackle this question I would be most grateful.

Thanks


Look in your formula booklet, you should see:

(1+x)n=1+nx+n(n−1)2x2+n(n−1)(n−2)3!x3+⋯\displaystyle (1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \cdots

To expand your formula, just replace all the places where you see an xx with x4\frac{x}{4}, so:

(1+x4)n=1+n(x4)+n(n−1)2(x4)2+n(n−1)(n−2)3!(x4)3+⋯\displaystyle \left(1 +\frac{x}{4}\right)^n = 1 + n\left(\frac{x}{4}\right) + \frac{n(n-1)}{2} \left(\frac{x}{4}\right)^2 + \frac{n(n-1)(n-2)}{3!} \left(\frac{x}{4}\right)^3 + \cdots
Original post by Zacken
Look in your formula booklet, you should see:

(1+x)n=1+nx+n(n−1)2x2+n(n−1)(n−2)3!x3+⋯\displaystyle (1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \cdots

To expand your formula, just replace all the places where you see an xx with x4\frac{x}{4}, so:

(1+x4)n=1+n(x4)+n(n−1)2(x4)2+n(n−1)(n−2)3!(x4)3+⋯\displaystyle \left(1 +\frac{x}{4}\right)^n = 1 + n\left(\frac{x}{4}\right) + \frac{n(n-1)}{2} \left(\frac{x}{4}\right)^2 + \frac{n(n-1)(n-2)}{3!} \left(\frac{x}{4}\right)^3 + \cdots


Wow, thanks again Zacken. Don't know what I would do without your wisdom.

So is there a chance of them asking to expand (2+X/4) then? Would the formula still apply? If so, would it be the same substitution?
Reply 3
Avatar for Zacken
Zacken
22
Original post by CrazyFool229
Wow, thanks again Zacken. Don't know what I would do without your wisdom.

So is there a chance of them asking to expand (2+X/4) then? Would the formula still apply? If so, would it be the same substitution?


Not quite, the substitution only works with (1+f(x))n(1+f(x))^n i.e: there needs to be a 1 in the first term of the bracket.

But there's a sneaky trick around this.

If you had (2+x4)n(2 + \frac{x}{4})^n - factor out a 2: (2(1+x8))n=2n(1+x8)n\bigg(2\left(1 + \frac{x}{8}\right)\bigg)^n = 2^n (1 + \frac{x}{8})^n.

Then you can just expand the latter fraction using the substitution of replacing wherever you see xx wuth x8\frac{x}{8} and then multiply all the terms of your series by 2n2^n.

i.e: (2+x4)n=2n(1+n(x8)+n(n−1)2(x8)2+n(n−1)(n−2)3!(x8)3+⋯)\displaystyle (2+ \frac{x}{4})^n = 2^n \bigg(1 + n\left(\frac{x}{8}\right) + \frac{n(n-1)}{2} \left(\frac{x}{8}\right)^2 + \frac{n(n-1)(n-2)}{3!} \left(\frac{x}{8}\right)^3 + \cdots\bigg)
Original post by Zacken
Not quite, the substitution only works with (1+f(x))n(1+f(x))^n i.e: there needs to be a 1 in the first term of the bracket.

But there's a sneaky trick around this.

If you had (2+x4)n(2 + \frac{x}{4})^n - factor out a 2: (2(1+x8))n=2n(1+x8)n\bigg(2\left(1 + \frac{x}{8}\right)\bigg)^n = 2^n (1 + \frac{x}{8})^n.

Then you can just expand the latter fraction using the substitution of replacing wherever you see xx wuth x8\frac{x}{8} and then multiply all the terms of your series by 2n2^n.

i.e: (2+x4)n=2n(1+n(x8)+n(n−1)2(x8)2+n(n−1)(n−2)3!(x8)3+⋯)\displaystyle (2+ \frac{x}{4})^n = 2^n \bigg(1 + n\left(\frac{x}{8}\right) + \frac{n(n-1)}{2} \left(\frac{x}{8}\right)^2 + \frac{n(n-1)(n-2)}{3!} \left(\frac{x}{8}\right)^3 + \cdots\bigg)


Thanks again!
I would rep you but it isn't letting me :frown:
Reply 5
Avatar for Zacken
Zacken
22
Original post by CrazyFool229
Thanks again!
I would rep you but it isn't letting me :frown:


Don't worry about it. :smile:
Reply 6
Avatar for Naughty1
Naughty1
1
Original post by Zacken
Look in your formula booklet, you should see:

(1+x)n=1+nx+n(n−1)2x2+n(n−1)(n−2)3!x3+⋯\displaystyle (1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \cdots

To expand your formula, just replace all the places where you see an xx with x4\frac{x}{4}, so:

(1+x4)n=1+n(x4)+n(n−1)2(x4)2+n(n−1)(n−2)3!(x4)3+⋯\displaystyle \left(1 +\frac{x}{4}\right)^n = 1 + n\left(\frac{x}{4}\right) + \frac{n(n-1)}{2} \left(\frac{x}{4}\right)^2 + \frac{n(n-1)(n-2)}{3!} \left(\frac{x}{4}\right)^3 + \cdots

Go to exam solutions
Reply 7
Avatar for Zacken
Zacken
22
Original post by Naughty1
Go to exam solutions


Compelling.
Original post by Zacken
Not quite, the substitution only works with (1+f(x))n(1+f(x))^n i.e: there needs to be a 1 in the first term of the bracket.

But there's a sneaky trick around this.

If you had (2+x4)n(2 + \frac{x}{4})^n - factor out a 2: (2(1+x8))n=2n(1+x8)n\bigg(2\left(1 + \frac{x}{8}\right)\bigg)^n = 2^n (1 + \frac{x}{8})^n.

Then you can just expand the latter fraction using the substitution of replacing wherever you see xx wuth x8\frac{x}{8} and then multiply all the terms of your series by 2n2^n.

i.e: (2+x4)n=2n(1+n(x8)+n(n−1)2(x8)2+n(n−1)(n−2)3!(x8)3+⋯)\displaystyle (2+ \frac{x}{4})^n = 2^n \bigg(1 + n\left(\frac{x}{8}\right) + \frac{n(n-1)}{2} \left(\frac{x}{8}\right)^2 + \frac{n(n-1)(n-2)}{3!} \left(\frac{x}{8}\right)^3 + \cdots\bigg)


shouldn't you be revising :smile:
Reply 9
Avatar for Zacken
Zacken
22
Original post by tinkerbella~
shouldn't you be revising :smile:


That's for nerds. :tongue:
Original post by Zacken
That's for nerds. :tongue:


you are a nerd :smile:
Original post by tinkerbella~
you are a nerd :smile:


Yes he is
Reply 12
Avatar for Zacken
Zacken
22
Original post by tinkerbella~
you are a nerd :smile:


Apparently not, since I'm not revising.
Original post by Zacken
Apparently not, since I'm not revising.


oh you got me there

Original post by Pablo Picasso
Yes he is


kinda cool for a nerd though
Reply 14
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Zacken
22
Original post by tinkerbella~
oh you got me there



kinda cool for a nerd though


'innit
Isnt this c4 not c2?
Reply 16
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Zacken
22
Original post by Mitul106
Isnt this c4 not c2?


Binomial expansion is C2 for Edexcel.

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