Sy123
This too shall pass
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- Nov 6, 2011
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- 2013
Holy shit I was that close to the actual solution. I got the integral part but from there everything went haywire.
In the future feel free to post your progress, in this case it was just a computation and the integral is not essential to understanding the problemHoly shit I was that close to the actual solution. I got the integral part but from there everything went haywire.
Perhaps an interesting excercise:
I'm not sure what I can argue that isn't calculus.
Try considering f(x+a) – f(x), where a is positive and x ≥ 1.I'm not sure what I can argue that isn't calculus.
Perhaps an interesting excercise:
That's as far as I could get. I know that it must be shown to be always true given the conditions y >= x > 1, but I'm not sure how.
I just realised it as I was posting. Great question!
yes please.oh nvm, just realised sy already posted solutions for the polygon and average distance question, no need to post those ones up then. If a 2016ers needs soln for the altitude q, I can type them up.
Which part do you specifically need help with(as in first part or second part)? I'll type up as soon as I come back from workyes please.
First...
second partWhich part do you specifically need help with(as in first part or second part)? I'll type up as soon as I come back from work
Alternatively, prove that if n is true, n+6 is true, then show trivially that it is true for the first six natural numbers and draw your conclusion from there.First...
Case 0: trivial, Case 1: LHS = 6, which is divisible by 6,
Case 2: LHS = 30, which is divisible by 6
Assume formula is true up to n=k:
k(k+1)(2k+1)=6M, M is a natural number
For n=k+1
LHS = (k+1)(k+2)(2k+3)
= (k+1)(2k^2 + 7k + 6)
= k(k+1)(2k + 1+ 6) + 6(k+1)
= k(k+1)(2k + 1) + 6k(k+1)+6(k+1)
= 6M + 6(k+1)^2
= 6K, K is a natural number
Since true for first case, and also for two consecutive cases, it is therefore true for all natural n
am I right to assume that it is asking for 12 is a factor of the n^4-n^2
Case n=0 or n=1
n^4-n^2 = 0, trivial
Case n=2
n^4-n^2 = 4*3 = 12, which is divisible by 12
Assume formula to be true, up to some natural number 'k'
k^4-k^2=12N where N is a natural number (not 0 or 1)
For n=k+1
LHS = (k+1)^4 - (k+1)^2
= k^4 + 4k^3 + 6k^2 + 4k+1 - k^2 - 2k -1
= 12N + 4k^3 + 6k^2 + 2k
= 12N + 2k(2k^2 + 3k+1)
= 12N + 2k(2k+1)(k+1)
It remains to show that n(n+1)(2n+1) is a multiple of 6, which was conducted earlier
= 12N + 2k*6Z (Z a natural number)
= 12(N+kZ)
Hence since the first case, two consecutive cases are true, the statement is proved true for all natural n.