induction questions that are hard to do in tha wee hours of tha mornin (1 Viewer)

[fashionista]

can any one help me with these induction questions which are realllly hard for me at 12.30 in the night??? damm brain switching offf after 7. yeh so here they r


n sigma r=1 (2r-1)^3=n^2((2n^2)-1)
annnnnnd heres the other one
show that [(x^n)-1] is divisible by x-1
thank u thank u
luv mee!!!!

 

[fashionista]

im sorry if the first question may be a bit hard to decipher but i cudnt think of ne other way of putting tha sigma in

 

[Xayma]

1st one.

1+27+125+...+(2n-1)^3=n^2(2(n^2)-1)

Let n=1

LHS=1
RHS=1=LHS
therefore, true for n=1

Assume true for n=k

1+27+125+...+(2k-1)^3=k^2(2(k^2)-1)

Prove true for n=k+1

1+27+125+...+(2k-1)^3+(2k+1)^3=k^2(2(k^2)-1)+(2k+1)^3

(Hmm gonna bash this as I cant see any little tricks)

RHS=2k^4-k^2+8k^3+12k^2+6k+1
=(k+1)^2(2{[k+1]^2} -1)
(might be an error or two in expanding but I didnt really factorise the bottom, I knew what it had to be)

therefore true for n=k+1 and since true for n=1 true for n=2,3.. and for all n>= 1

 

[fashionista]

thank u!! can i jus ask plz how u got k^2(2(k^2)-1)+(2k+1)^3??? wen u were proving tru for n=k+1?? wen i put k+1 in for n i got a totally diff answer and i cant seem to get y u got this answer.
thanx!!!

 

[Xayma]

As you prove for n=k+1 you add (2k+1)^3 to the left hand side therefore it also must equal the right hand side plus (2k+1)^3. If you just input k+1 for k in the right hand side you get nowhere. you are trying to achieve an answer where k+1 exists fork. *Prays that makes sense*

 

[fashionista]

OOOOOHHHHH!!!! i getcha now!! cuz for all my other induction question it worked wen i put n=k+1 in for where there was any n's but i gess it doesnt work for all of em. thank u thank u thank u (it did make sense)

 

[KeypadSDM]

For number 2:

R.T.P (x - 1) factorises (x^n - 1)

For n = 1
(x^1 - 1) = 1 * (x - 1)
:. True for n = 1

Assume true for n = k

:. x^k - 1 = Z(x) * (x - 1)
(Where Z(x) is a polynomial in x where the leading co-efficient is x^(k-1))

For n = k + 1

x^(k + 1) - 1
= x^(k + 1) - x + x - 1
= x(x^k - 1) + (x - 1)
= Z(x) * x * (x - 1) + (x - 1)
= (x - 1) * (x *Z(x) + 1)

:. True for n = k + 1 if true for n = k
:. True for all n >= 1 (n E Z)

 

[Grey Council]

whoa

whoa, keypad, you have complicated it for no reason. It can be done a much simpler way.

Anyway, as for the 1st one, like Xayma got:
2k^4 + 8k^3 +11k^2 + 6k + 1

I have also managed to get this. However, as we have to prove the this is equal to (k+1)^2(2k^2 + 4k + 1) we are not allowed to use this fact. We can not use what we know the answer is going to be, can we? I mean, i could expand out the answer, and get it to equal 2k^4 + 8k^3 +11k^2 + 6k + 1 but we aren't allowed to do this.

So my question is this: how do we factorise:
2k^4 + 8k^3 +11k^2 + 6k + 1
to get
(k+1)^2(2k^2 + 4k + 1) ?

 

[mazza_728]

For question 1:
ive got LHS = 2k^4+8k^3+11k^2+6k+1
But then when i totally expand RHS I get
RHS=2k^4+12k^3+11k^2+6k+1
I've checked this and i keep getting the same answer. can one of uz try totally expanding both sides please?

 

[mazza_728]

and i do question 2 a diff way to u guys:
Prove x^n - 1 is divisible by x-1
Show true for n=1
x-1=x-1
Assume true for n=k
x^k - 1 = (x-1)M where M is an integer
x^k = ((x-1)M) + 1
Prove true for n=k+1
x^(k+1) - 1
=(x^k (mutiplied by) x) -1
=(((x-1)M+1)x)-1
If anyone is still following me? well im lost there! i cant get any further!! can anyone help?
If i expand this i get
=mx^2-mx+x-1
How do i factor this to somehow get (x-1)?
Yeah i dont get it??/
Anyone else?

 

[Grey Council]

=mx^2-mx+x-1
How do i factor this to somehow get (x-1)?

lol, this is rather simple:
Take mx from the first two variables:
= mx(x - 1) + x - 1
= (x - 1)(mx + 1)
which is divisible by x - 1.

hehe, glad to be of help, now go and help me answer the question i asked.

 

[ND]

Re: whoa

Originally posted by GuardiaN
whoa, keypad, you have complicated it for no reason. It can be done a much simpler way.

How so? keypad's and mazza's methods are basically the same, only that keypad's would probably be a little quicker cos you didn't have to bother expanding that thing.

Anyway, as for the 1st one, like Xayma got:
2k^4 + 8k^3 +11k^2 + 6k + 1

I have also managed to get this. However, as we have to prove the this is equal to (k+1)^2(2k^2 + 4k + 1) we are not allowed to use this fact. We can not use what we know the answer is going to be, can we? I mean, i could expand out the answer, and get it to equal (k+1)^2(2k^2 + 4k + 1) but we aren't allowed to do this.

So my question is this: how do we factorise:
2k^4 + 8k^3 +11k^2 + 6k + 1
to get
(k+1)^2(2k^2 + 4k + 1) ?

It really doesn't matter; you can just say:

"For it to be true for n=k+1, RHS=(k+1)^2(2k^2 + 4k + 1)=2k^4 + 8k^3 +11k^2 + 6k + 1"

Then start with LHS and proceed to prove it.

 

[Xayma]

Originally posted by ND
It really doesn't matter; you can just say:

"For it to be true for n=k+1, RHS=(k+1)^2(2k^2 + 4k + 1)=2k^4 + 8k^3 +11k^2 + 6k + 1"

Then start with LHS and proceed to prove it.

Yeah I know, our teacher just says not to do it to prevent getting stuck but I prefer it that way. Cause it shows you know what youre trying to get.

 

[KeypadSDM]

Originally posted by mazza_728
x^k - 1 = (x-1)M where M is an integer

Dude ... M is a polynomial in x, with the leading co-efficient equal to x^(k-1)

It's only an integer for k = 1

Strangely enough, this method is identical to mine.

I'm pretty sure it's the quickest method via induction. But is an alternate proof available due to the remainder theorem?:

Let f(x) = x^k - 1

:. f(1) = 1^k - 1
= 0

:. (x - 1) is a factor of (x^k - 1)

Is that valid?

 

[Xayma]

Since k must be a positive integer I cant see why it wouldnt be.

 

[mazza_728]

Originally posted by mazza_728
For question 1:
ive got LHS = 2k^4+8k^3+11k^2+6k+1
But then when i totally expand RHS I get
RHS=2k^4+12k^3+11k^2+6k+1
I've checked this and i keep getting the same answer. can one of uz try totally expanding both sides please?

Did any of you guys try this?

 

[evilc]

i just did it and will (attempt to) attatch my solution

 

[evilc]

here is my solution...the concluding statement is cut off though

 

[mazza_728]

hey ... got it thankyou .. just another one of my stupid mistakes!
ta
xoxo

 

[Grey Council]

Originally posted by ND
It really doesn't matter; you can just say:

"For it to be true for n=k+1, RHS=(k+1)^2(2k^2 + 4k + 1)=2k^4 + 8k^3 +11k^2 + 6k + 1"

Then start with LHS and proceed to prove it.

I was thinking that, wasnt too sure though. Like in circle geometry or trig. hehe, its the only way i could think of doing it.

Originally posted by KeypadSDM

I'm pretty sure it's the quickest method via induction. But is an alternate proof available due to the remainder theorem?:

Let f(x) = x^k - 1

:. f(1) = 1^k - 1
= 0

:. (x - 1) is a factor of (x^k - 1)

Is that valid?

o! No wonder you topped the state, I didnt even think about that.

But yeah, I can't see why it wouldnt be accepted, unless the question asked you to prove via induction.

And this is how i'd do it:
R.T.P (x - 1) factorises (x^n - 1)

For n = 1
(x^1 - 1) = 1 * (x - 1)
:. True for n = 1

Assume true for n = k
ie (x^k - 1) = M * ( x - 1 )

Prove true for n = k + 1
x^(k+1) - 1
= x.x^k - 1
= x(x^k - 1 ) + x - 1
= x(m(x-1)) + x - 1
= (x-1)(xm + 1)

I thought Keypad's method a bit too complicated cause i've never seen his way. He is on a higher plane of maths existence ( ) than me, so I thought he just thinks a bit too complicated. Im prolly wrong, but thats cause i didnt quite get what the hell he was saying in his proof. hehe, my bad.