Can you assume this identity? (1 Viewer)

[~ ReNcH ~]

<b>(1+x)ⁿ(1+x)ⁿ = (1+x)²ⁿ</b>

In past trial and HSC papers, they've stated it at the beginning of the question. If they <b>don't</b> state it, can you just assume that it's true and then go ahead and use it to answer the question?

 

[Rorix]

man i dunnoooooooooo


its a bit tuff to prove



yes, i'm joking

 

[Estel]

<b>(1+x)ⁿ(1+x)ⁿ = (1+x)²ⁿ</b>

In past trial and HSC papers, they've stated it at the beginning of the question. If they <b>don't</b> state it, can you just assume that it's true and then go ahead and use it to answer the question?

I'd have to disagree with Rorix on this one.... (I'm running the assumption you're being sarcastic... and if I'm wrong....)

Are we allowed to assume 1 + 1 = 2.... I've always thought calling things axioms was a bit dodgy...

Now for a challenge question: prove it.

 

[ae]

offcoz u can assume tat
its same as saying y^1 X y^1 = y^2

 

[~*HSC 4 life*~]

<b>(1+x)ⁿ(1+x)ⁿ = (1+x)²ⁿ</b>

In past trial and HSC papers, they've stated it at the beginning of the question. If they <b>don't</b> state it, can you just assume that it's true and then go ahead and use it to answer the question?

binomial questions right?

well they usually lead you into the proof, and i guess you can use it because it's just indicie rules? not sure, i suck @ binomials

 

[Rorix]

I'd have to disagree with Rorix on this one.... (I'm running the assumption you're being sarcastic... and if I'm wrong....)

highlight the text below my frown

 

[ngai]

nah, u gotta prove it by expanding each brackets using the binomial theorem
coz unless they say its given, then theyre testing to see if u know what binomial theorem is

 

[breaking]

yeah, what she ngai said...

 

[Rorix]

nah, u gotta prove it by expanding each brackets using the binomial theorem
coz unless they say its given, then theyre testing to see if u know what binomial theorem is

but i don't know what n is ????

like, i tried subbing like n= 3.5

but i couldn't do (7/2)C0 on my calculator?

maybe i need to get a new model

 

[withoutaface]

let m=1+x

mⁿ*mⁿ
m²ⁿ

surely you don't need any more than that.

 

[Rorix]

*scratches head*

are you sure let m=1+x is allowed?

like ngai said im pretty sure you have to expand....like
LHS= ((n 0) + (n 1)x + (n 2)x^2.....+(n n)x^n)^2

......

 

[mervvyn]

I like what withoutaface did... i can't see why they would need you to prove something like that at all :S

That, and it would be a pain in the arse to expand that. It's not like time grows on trees in exams.

 

[Rorix]

but what if x=-2 and n=1/2?

 

[Xayma]

Well then (1+x)ⁿ isn't real.

And the other method would still be invalid.

 

[Rorix]

I am rolling my eyes at this thread over the internet

 

[Slidey]

But what if you're not allowed to roll your eyes over the internet?

 

[Rorix]

on my internet we obey my rules

 

[~ ReNcH ~]

let m=1+x

mⁿ*mⁿ
m²ⁿ

surely you don't need any more than that.

So that's all you need to include?

I'm getting a bit confused now.

 

[Estel]

I'm still at a loss as to how you would "prove" the result.

It's a bit like asking someone to "prove" 2.2 = 2²

 

[withoutaface]

I'm still at a loss as to how you would "prove" the result.

It's a bit like asking someone to "prove" 2.2 = 2²

It IS asking someone to prove 2.2=2²