Sum each one individually:
For Question Pell:
Your first value is wrong -> which has made the whole sequence wrong.For Question Pell:
i) 1, 1, 3, 7, 17
Sorry, not familiar with induction...
Wait, we actually count the zero as the first one?Your first value is wrong -> which has made the whole sequence wrong.
Correct, but those are the easy marks :sWait, we actually count the zero as the first one?
If so, then yeah it should be: 0, 1, 2, 5, 12
Correct - Nice work. I am working on a Complex Numbers question at the moment, should be good.Doing them on paper, won't TeX up every step.
i) 0, 1, 2, 5, 12
ii)
iii) Subbing in x,y and expanding, you get the RHS from part ii. Now because you're using n=2k and k is integral, (-1)^(2k) = 1 as required.
iv) Sub in the given expressions for x and y, then sub in m2 = k(k+1)/2. It cancels to equal 1.
v) The desired equation can be reached by equating the two expressions given for y to get an expression for m, then squaring this to get Np. However, I'm reluctant to do this, because the fact that a and b both satisfy an equation does not mean that a = b. I think my working would be sufficient in an exam, but I'd need to justify this step to myself properly.
vi) By manually computing a few more Pell numbers, It's easy to convince oneself that the ratio between consecutive terms converges to 1 + sqrt(2) (not very mathematical, hehe). Now expanding the limit we're meant to find, we get (P2p+2 / P2p)2. The nice limit equalit means that this ratio by itself is (1+sqrt(2))^2 -- squaring this again gives (1+sqrt(2))^4 which is the answer.
For the first one I got 425Here's a tricky little perms & combs for y'all.
Surely it's more than that though.For the first one I got 425
The figurate numbers give the triangular numbers, sum of tri numbers (tetra), sum of that, and so on:Using triangular numbers I ended up with:
But this was way too tedious to evaluate.
Is my reasoning correct?The answer is 4 digits long. I've verified it. It shouldn't be tedious to compute -- I did it quickly using only the operations available on a standard BOS-approved calculator.
Part ii) perhaps requires slightly more logical reasoning than you'd get in an HSC paper, but part i) is about the same difficulty as the perms+combs HSC questions of past years.
It looks like you have some good ideas, but it needs refining. For your case 3, for example, the correct answer is 35, but your method gives 30. I think it's because of this: "we can pick Case 2 becasue we just add on one fruit to each permutation from Case 2". But case 2 includes case 1, so when you add one fruit to one of case 2's permutations that it borrowed from case 1, you'll now have one fruit picked 3 times, which will be double-counted when you separately add on case 1. Also, you'd need to account for which one of the fruit in the pair that you add one to -- but again, you can't just multiply by 2, because that screws up the permutations case 2 borrowed from case 1 where two of the same fruit were picked.Is my reasoning correct?
I need to get better at perms and combs
Sounds goodLet us establish the fact that if we pick an odd number of fruits, we cannot retain proportionality.
Hm ok... but what about the arrangement (3 grapes, 3 apples)? This is an arrangement for 6 fruits, and it's not proportional to any arrangement of 3 fruits , but yet it isn't legal...Now consider picking 6 fruits (we doubled it), now we can definitely say that proportionality of the ways to pick 3 fruits remain.
Hence the new total permuations for 6 fruits chosen is
(Arrangements for 6 normally) - 35