A Question from my first assessments (1 Viewer)

[Grey Council]

A few Questions from my first assessment (SBHS 04)

Prove, by mathematical induction:

(1^5 + 2^5 + 3^5 +..+ n^5) + (1^7 + 2^7 + 3^7 +...+ n^7) = 2(n/2(1+2+3+..+n))^4

Looks ugly, nie? Our teachers (not just mine ) said that this question is the HARDEST (read: one of the hardest) that they can ask you in four unit.

I didn't get it in the exam, I got scared off. I looked at it, and I thought why waste time. Go do some of the later questions. And it was a wise decision too.

Anyhow, I'd like to see people's solutions. I'll more likely than not learn a new technique from Keypad/ND.

 

[KeypadSDM]

Shouldn't this be in the 4 unit section?

 

[KeypadSDM]

Not to be picky, but I don't think it works.

For n = 1

RHS = 1/8, LHS = 2, doesn't it?

 

[KeypadSDM]

But then again it works for n = 2... Maybe i'll give this a go then.

 

[ND]

Here's something that may help:

(1^5 + 2^5 + 3^5 +..+ n^5) + (1^7 + 2^7 + 3^7 +...+ n^7)
= 1^5(1+1^2) + 2^5(1+2^2) + ... + n^5(1+n^2)

and also:

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

so it comes down to proving:

1^5(1+1^2) + 2^5(1+2^2) + ... + n^5(1+n^2) = 2((n^2/4)(n+1))^4

which doesn't look as complicated. Though it could still prove difficult.

 

[Grey Council]

its a 3u question.

and the RHS is (this isn't too clear) : ((n^4)/8)(1+2+3+..+n)^4

ie, thats the RHS expanded.

And i KNOW how to do it, i was just wanting to see how you guys did it. I want to learn new techniques.


And it isn't THAT complicated. I think you'd be better off writing it out on paper.

 

[KeypadSDM]

Nonetheless, you should have told us the limit of n>=2

 

[ND]

Umm i don't think that it's true; for n=3 LHS=2592 and RHS=13122.

 

[KeypadSDM]

Lol, we need it written down.

 

[Grey Council]

ayo! I'll see if I can write it down a bit more clearly:

Prove by mathematical Induction, that for all positive integers n the LHS is equal to the RHS, where:

LHS=
(1^5 + 2^5 + 3^5 + ... + n^5) + (1^7 + 2^7 + 3^7 + ... + n^7)

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

O shite, I thought I knew how to do it, now im not so sure. Darn.

EDIT:
Actually, on further thought, my method is ok.

 

[Grey Council]

A second question:
APB is a semi circle centre O, touching three sides of a rectangele ABCD.

AMPC is a straight line, a diagonal of the rectangle ABCD. M is the midpoint of AP.

Prove that:
i) O, M, C and B are concyclic.
ii) <AMB = 135.

Hope you can draw the diagram. Shouldnt be too hard. Draw a rectangle ABCD such that its width is twice its height, put a semicircle in it, with the diameter as AB (its width) and radius as its height, ie, it touches the DC. Then go from there.

 

[ND]

For the induction question:

(1^5 + 2^5 + 3^5 +..+ n^5) + (1^7 + 2^7 + 3^7 +...+ n^7)
= 1^5(1+1^2) + 2^5(1+2^2) + ... + n^5(1+n^2)

2(1 + 2 + 3 + ... + n)^4 = 1/8*(n(n+1))^4

so we must prove:

1^5(1+1^2) + 2^5(1+2^2) + ... + n^5(1+n^2) = 1/8*(n(n+1))^4

for n=k:

1^5(1+1^2) + 2^5(1+2^2) + ... + k^5(1+k^2) = 1/8*(k(k+1))^4

now for n=k+1:

LHS= 1^5(1+1^2) + 2^5(1+2^2) + ... + k^5(1+k^2) + (k+1)^5(1+(k+1)^2)
= 1/8*(k(k+1))^4 + (1+(k+1)^2)
= 1/8*(k+1)^4(k^4+8k^3+24k^2+32K+16) (upon factorising and expanding)
=1/8(k+1)^4((k+1)+1)^4

.'. it's true for n=k+1

quite straight forward.

For the geometry:

AB=2BC
/_APB=90* (angle in semi-circle is right)
/\AMO and /\APB are similar (i'm not gonna go through proving their similar as it's really easy)
/_AMO=/_APB (corresponding sides in similar triangles are equal)
.'. /_AMO=90*
/_CMO=90* (180* straight line)
.'. OMAC is a cyclic quad (opp angles are supplementary)

edit: oops i missed part ii) (when you do it, leave a space between the < and the angle so it doesn't hide it.)

/_OCB=45* (as OB=BC)
/_OCB=/_OMB (angles in the same segment are equal)
/_AMB=/_AMO+/_OMB
= 90*+45*
=135*

 

[Grey Council]

oo k. I'll post up my method later on today. I hafta go now. But in the exam, the first one scared the shit outta me, and i didn't even attempt it.

For the second one, I got the first part, not the second one. ;(

 

[KeypadSDM]

Originally posted by GuardiaN
oo k. I'll post up my method later on today. I hafta go now. But in the exam, the first one scared the shit outta me, and i didn't even attempt it.

For the second one, I got the first part, not the second one. ;(

When you get the the HSC, don't let anything scare you. If you've done enough work, you'll have a look at it later in the exam, and you'll see that it's simpler than it initially appears.

 

[Grey Council]

true true. Thing is, the teacher who set our exam is famous for finding questions from foreign abstract mathematical books, and as soon as i saw the n^5's and the n^7's I freaked out. lol, i wrote down the steps till: prove for n = k+1. teehee, i got 2 marks out of the 6. ANYWAY, here is how i did it

prove true for n=1
assume true for n=k
prove for n=k+1:

Sum of k+1 terms = sum of k terms + the k+1'th term

therefore:
1/8*(k(k+1))^4 + (k+1)^5 + (k+1)^7
has to equal to
1/8*((k+1)(k + 2))^4

then you make the denominator of the the whole first term = 8. Factorise, and you get:
1/8*(k+1)^4(k^4+8k^3+24k^2+32K+16)
using pascal's triangle, this is (k+2)^4

Ofcourse I wrote it down with full explanation while i was doing it, more the help myself remember (this was after the exam) what i was on about. heheh, hope it makes sense.

ooh, btw, can you say that:
(1^5 + 2^5 + 3^5 + ... + n^5) = n/2*(1 + n^5)
?

You can't, can you? its not a AP. ah well, for some reason or other, I thought i could say that. then today i realised I couldnt.

 

[ND]

Yeh that solution is exactly the same as mine, only i factorised the LHS in anticipation of a more difficult question.

Originally posted by GuardiaN
ooh, btw, can you say that:
(1^5 + 2^5 + 3^5 + ... + n^5) = n/2*(1 + n^5)
?

You can't, can you? its not a AP. ah well, for some reason or other, I thought i could say that. then today i realised I couldnt.

Nope that doesn't work, as you said, it's not an AP. (the difference between the terms k and k+1 is not the same as between k+1 and k+2)

 

[Grey Council]

hehe, as i said, not that difficult. You factorised the LHS and i was like, wtf he do? lol

Well done. If only i hadn't been scared off in the exam.

 

[Grey Council]

o, sorry, just read the fact that you guys didn't recieve the second part of the second question. Hang on, here it is:
prove angle AMB = 135

There you go. Its easier than the first part, imho.

 

[ND]

Originally posted by GuardiaN
o, sorry, just read the fact that you guys didn't recieve the second part of the second question. Hang on, here it is:
prove angle AMB = 135

There you go. Its easier than the first part, imho.

I posted it already:

edit: oops i missed part ii) (when you do it, leave a space between the < and the angle so it doesn't hide it.)

/_OCB=45* (as OB=BC)
/_OCB=/_OMB (angles in the same segment are equal)
/_AMB=/_AMO+/_OMB
= 90*+45*
=135*

 

[Grey Council]

o, how the hell did you know the question? wtf :S