i just did when n = 2 and 3
That's goodi just did when n = 2 and 3
and using the solution function, i got my answer right for n=2 and 3
Just curious, could you do the question I posted?That's good
Even though I have seen it before, trying it when you put it up I got to a point but after that I wasn't sure, especially due to my lack of understanding of a lot of notation (only recently knew about double factorial, sad I know), and I'm quite bad with number theory and quite oblivious when it comes to proving things.
Start with some easier ones then, instead of trying hard ones straight away. For example if you couldn't simplify the expressions, go over the basics from a textbook then comeback and try.
haha I haven't solved it yet either, I got up to this, but realised my approach probably wouldn't work that well:Even though I have seen it before, trying it when you put it up I got to a point but after that I wasn't sure, especially due to my lack of understanding of a lot of notation (only recently knew about double factorial, sad I know), and I'm quite bad with number theory and quite oblivious when it comes to proving things.
Short answer, nope
lol I thought this was a year 9-10 questionhaha I haven't solved it yet either, I got up to this, but realised my approach probably wouldn't work that well:
I'm going to try another approach though and see how that works out.
lol m8haha I haven't solved it yet either, I got up to this, but realised my approach probably wouldn't work that well:
I'm going to try another approach though and see how that works out.
I beg to differ. 'Ymcaec' is on the case.lol m8
if you haven't solved it yet
it's highly unlikely a year 9/10 will
(someone pls feel free to prove me wrong)
Someone other than him, he's already acceleratedI beg to differ. 'Ymcaec' is on the case.
for i)Another question for you guys
Edit: arrgh, latex code doesn't seem to work
Let me know what you decide to do. I might have another look soon.haha I haven't solved it yet either, I got up to this, but realised my approach probably wouldn't work that well:
I'm going to try another approach though and see how that works out.
It uses logic and skills most Year 10s have though at quite an advanced level.lol I thought this was a year 9-10 question
O_O this is from cambridge right?Another question for you guys
Edit: arrgh, latex code doesn't seem to work
Another question for you guys
Edit: arrgh, latex code doesn't seem to work
Wow! Nicee That methodology was very similar to the question I was talking about lol}\\=\frac{1}{n(n+1)}\\\\\\$ii. If n = 1, $n(n+1) = 1\times2 \\$Hence $(\frac{1}{1} - \frac{1}{1+1}) +(\frac{1}{2} - \frac{1}{2+1}) +(\frac{1}{3} - \frac{1}{3+1}) + ... + (\frac{1}{100} - \frac{1}{100+1}) \\\frac{1}{1}+(\frac{1}{2} +\frac{1}{3} + ... + \frac{1}{100}) - (\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{100}) - \frac{1}{101}\\=\1 - \frac{1}{101}\\ =\frac{100}{101})
