Stuck on some algebra for MI (1 Viewer)

[DistantCube]

how on earth do you get;

(k+1)^3 + k^2/4 . (k+1)^2

to equal:

(k+1)^2 /4 . (k+2)^2

EDIT: if you can't read that; then:

     [SIZE=1]3[/SIZE]     [SIZE=1]2[/SIZE]         [SIZE=1]2[/SIZE]
(k+1)  +  [u]k[/u]  .  (k+1)
          4


=

     [SIZE=1]2[/SIZE]         [SIZE=1]2[/SIZE]
[u](k+1)[/u]  .  (k+2)
 4

 

[Estel]

(k+1)^3 + k^2(k+2)^2/4 - [(k+1)^2(k+2)^2/4]
= (k+1)^2(4k+4-k^2-4k-4)/4 + k^2(k+2)^2/4
= -k^2(k+1)^2/4 + k^2(k+2)^2/4 =/0 ---> error. your statement is not true.

 

[DistantCube]

no, I know they're equal, but its MI, I have to work the LHS out to look exactly like the RHS...like trig identities...I don't think you can simply put them on the one side and show that it equals 0...

As in, I've only shown you guys what its supposed to look like, which is the (k+1)^2/4 . (k+2)^2 part, you cant simply equate and show they're equal, you must use the assumption and prove the (k+1)th term, so you have to manipulate till the first one looks like what it should look like...

 

[Slidey]

I haven't read too far, but always remember this:

If you need to PROVE that LHS=RHS, then showing that LHS-RHS=0 is sufficient proof.

 

[DistantCube]

read further...its not simply a matter of them giving you LHS and RHS then equating etc; you have deduced LHS, you know what RHS *SHOULD* be, so you manipulate LHS till it looks like RHS...ffs, has no one done mathematical induction!?!

edit: My apologies, I don't mean to sound like an asshole, but this is annoying me, I think I'll just leave it be and ask my teacher tomorrow if no one knows.

 

[Estel]

see above: I thought your statement was true so when I got to the step of subtracting, I didn't bother to check that the terms were the same.
In fact, your statement isn't true. I threw it into the graph calculator to confirm that.

Pass the original induction q, or do it yourself again.

 

[Slidey]

So you're saying: You have LHS. You have an idea what the RHS is, but want to prove that LHS is equal to the RHS you think it equals?

If so, then proving LHS-RHS=0 is sufficient proof.

I taught myself induction last night and it involved proving LHS=RHS. I'm pretty sure that's what you're trying to do here.

Look, post up the entire induction question for me, OK?

 

[Slidey]

In fact, your statement isn't true. I threw it into the graph calculator to confirm that.

Good. Restores some confidence.

 

[DistantCube]

Estel....no....

Slide, good idea, it's probably my fault, should have asked the complete question, I have a worked answer but one line is tripping me out, I think i know WHAT theyve done but no clue HOW they've done it;

QUESTION:
Use Mathematical Induction to prove:

1^3 + 2^3 + 3^3 + ... + n^3 = n^2/4 . (n+1)^2

------------Worked Sol'n.-------------

let Sn[/2ize]= 1^3 + 2^3 + ... + n^3
RTP Sn[/2ize]= n^2/4 . (n+1)^2

For n=1, LHS =1, RHS = 1/4 .4 = 1, so statement is true for n=1

Assume true for n=k; i.e: k^2/4 . (k+1)^2

Show true for n=(k+1);

Sk[/2ize] = k^2/4 . (k+1)^2
Tk+1[/2ize] = (k+1)^3

therefore; Sk+1[/2ize] = Sk[/2ize] + Tk+1[/2ize]
=k^2/4 . (k+1)^2 + (k+1)^3
=1/4 . (k+1)^2 [k^2+4(k+1)] *****************
=1/4 . (k+1)^2(k+2)^2
=(k+1)^2/4 . (k+2)^2

so the statement is true for n=k+1, hence by MI it is true for n>=1

--------------------------------

How did they do the line that I put all of the asterix by, that's all I don't get, I *think* I know what they did, but unsure how...how did they replace the k^2 and get rid of (k+1)^3 to be able to multiply [k^2+4(k+1)]??

 

[Slidey]

1^3 + 2^3 + 3^3 + ... + n^3 = n^2/4 . (n+1)^2

Test for n=1:
LHS=1=RHS
Assume true for n=k:
1^3 + 2^3 + 3^3 + ... + k^3 = k^2/4 . (k+1)^2
Test for n=k+1:
1^3 + 2^3 + 3^3 + ... + k^3 + (k+1)^3 = (k+1)^2/4 . (k+2)^2
LHS=k^2/4 . (k+1)^2 + (k+1)^3
=(k+1)^2(4k+4+k^2)/4
=(k+1)^2.(k+2)^2/4
RHS=(k+1)^2/4 . (k+2)^2
=(k+1)^2.(k+2)^2/4
=LHS
Therefore true for k+1, but since it is true for n=1, then it is also true for n=2, n=3, et cetera. It is true for all natural n.

 

[DistantCube]

k^2/4 . (k+1)^2 + (k+1)^3
=(k+1)^2.k^2/4 + (k+1)^2.4(k+1)/4
=(k+1)^2.k^2/4 + (k+1)^2.(4k+4)/4
=(k+1)^2.(k^2+4k+4)/4


You have got to be kidding me, that's it....lol, can't believe that was it. *sigh* I need rest. Thanks slide.

 

[Estel]

look at your original post and see how it's actually different to k^2/4 . (k+1)^2 + (k+1)^3