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I agree, permutation and combinations is hard! Even at uni I did a subject called Combinatorial Probability and it was one of the toughest subjects of the whole Actuarial Studies degree! I also agree that nothing beats practicing the broadest range of questions possible. These questions don't...

Yes good point. Please disregard my comments 😅

For Q1, I think an alternative approach could be to use roots of unity. For example, consider the solutions to z^3 - 1 = 0 ... Plotting these in the unit circle would give rise to an equilateral triangle, each solution would have modulus of 1, and the sum of the roots would add up to 0 ... For...

Legendary. Thank you!

Hi all, I'm a bit stuck on part ii) of this question. Part i) was easy enough, and I also know how to prove part ii) using Pascal's identity. But I can't see the link between part i) and part ii), which seems to be implied by the word "hence" in the question. Any ideas?

I've also recorded a video of my working in case you learn better that way:

Hi there. I didn't do Standard so haven't technically studied the networks/critical path topics. But I'd be willing to try and get my head around it if you haven't had anyone else able to help yet. Perhaps you could send through one question you are struggling with at www.tickboom.study, and I...

Nice one. It's a great question!

Brilliant! Thanks so much!

Does anybody have any ideas for part ii)? I can show it for n=3, but can't work out how to show it in general. I tried induction but it didn't seem to work for me.

Yes you're right. I misspoke when I referred to that as the binomial expansion. I should have said "binomial series" which is a power series that I definitely did not learn until university.

I finally found the time to work through this, and yes I was also able to work it out by applying the Stolz-Cesaro theorem twice. So glad to have finally solved this! For anyone interested in how to tackle this, you can see my working here:

Wow thanks so much! I will get Googling on the Stolz–Cesàro theorem and see if I can work it out. Agreed, I think this question is a beyond the syllabus.

Yes that's a good point. I tried taking that path, and I used the technique where you state Sn+1 in terms of Sn, substitute C for both Sn+1 and Sn, and then solve for C, but unfortunately it didn't lead anywhere.

I've been doing some research on the monotone convergence theorem, and as far as I can understand, it is focused on the convergence of a sequence (i.e. showing that the limit of the values in a sequence is some value). However, in this question I think we are more focused on the convergence of...

I never had any tutoring and ended up doing pretty well, particularly in maths (99 for 3U and 94 for 4U, with a UAI of 98.75). That said, in hindsight I think tutoring for 4U would have helped because I think the content can get beyond a lot of teachers, so many 4U students can feel like they...

Good luck to everyone sitting their math exams next week! I thought I'd share my top 3 tips for heading into math exams (and probably any other exam too) ...

Is there a little ambiguity here? Does increasing order mean always increasing by 1? For example, 12345, 23456, 34567, 45678 and 56789. Or could it also include examples like 13456, 14567, 15678, 16789?

I've copied Q16 below. From what I can see, the denominator in the sigma term is 2k+1 (highlighted yellow). Do you have something different?