A Little Gem (1 Viewer)

[yrtherenonames]

Just thought I'd share the love from our 3unit Trial; most of it was fine except for the incomprehensible final question. I suspect it involves factorials. To quote

"Use induction, or otherwise, to prove that the sum of the products of all the pairs of different integers that can be formed from the first n positive integers is
n(n-1)(n+1)(3n+2)/24
Enjoy!

 

[David_O]

Hate to fluff it for you, but we got that on our first induction hw sheet.

n(n-1)(n+1)(3n+2)/24 + (n+1)(1+2+3...+n) using assumption
= n(n-1)(n+1)(3n+2)/24 + n(n+1)^2/2
= n(n+1)[(n-1)(3n+2) +12(n+1)]/24
= n(n+1)[3n^2 +11n +10]/24
= n(n+1)(n+2)(3n+5)/24, as required.

 

[yrtherenonames]

Well David O, if that is your real name, that information would've been more useful yesterday!!
Jks; it's nice that its easy but i had no friggin idea what it was talking about.

 

[shafqat]

Hope it was easier than last years trials. Everyone screwed that one up.

 

[yrtherenonames]

It was fair, I thought.
Your Trial Question 7 was a biatch; our one had
a)3-D trig
b)A doable binomial proof
c)Mother of a Question
4 Marks each.

 

[withoutaface]

Use telescopic sums:

the question is essentially asking you to find:

(1+2+...+n)² - (n²+...+2²+1</sup>2</sup>)


Find formulae:
Let S = 1+2+...+n
S = n+...+2+1
2S = (n+1)+...+(n+1)+(n+1)=n(n+1)
S = n(n+1)/2
1+2+...+n = n(n+1)/2

1²+2²+...+n²=n(2n+1)(n+1)/6
(Proof for this can be found at http://www.maths.usyd.edu.au/u/UG/JM/MATH1903/r/notes1.pdf, page 8)


:. the answer is:
(n(n+1)/2)²-n(2n+1)(n+1)/6
n²(n²+2n+1)/4 - n(2n²+3n+1)/6
(6n⁴+12n³+6n²-8n³-12n²-4n)/24
(6n⁴+4n³-6n²-4n)/24
2n(3n³+2n²-3n-2)/24
n(n+1)(n-1)(3n+2)/12

I suspect I've made a carrying error somewhere

Without giving you a hint about telescopic sums this seems to be an overly complicated problem for a 3u test though:S

 

[danza108]

if it isnt too much trouble, i wonder if you post up the grammar paper, i always like doing them, damn hard tho

 

[yrtherenonames]

Only the very last question was genuinely hard; I'll try and post it tomorrow.