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We have assumed that the statement is true for n=k (1+2+4+...+2(k-1) = 2k - 1). To prove true for n=k+1, you then have to add 2(k-1+1), i.e. 2k, to the left hand side, since the statement is a sum of terms and the (k+1)th term is 2(k-1+1). I_F

The problem with arguing about religion is that the very basis of religion is a belief in something you can't understand or comprehend in your lifetime. Part of the problem is that neither side will listen to the other and both sides regard themselves as above the other. Get rid of that attitude...

Ranks are important, but they're not the be-all and end all. It's better to be last in a class that averages 90 in the HSC exam than to come first in a class that averages 50 in the HSC exam. If your rank is relatively low, but your class is very competitive, then you shoulodn't be too...

With the average at 70, I'd keep it. Remember the 4-unit guys don't actually complete the 2-unit course, and the 3-unit people are expected to be ahead of the 2-unit guys. I_F

No problemo. :) I_F

This is a geometric series with common ratio 2, since the variable is in the exponent. However, if you are asked to prove through mathematical induction, then you cannot use the series formula. You must use the induction process or you get nothing. :) BTW for arithmetic series you use T...

The best advice I can give is to run the Connection Wizard (if you haven't already), found by clicking "Set up a Home or Office Connection" in the "Network Tasks" panel in Network Connections. Run it once on the desktop, setting it to "This computer connects directly to the internet", and then...

1+2+4+...+2(n-1) = 2n - 1 Step 1. Prove for n = 1 (smallest n) LHS = 1 RHS = 1 I.e. true for smallest n. Step 2. Assume true for n = k, i.e. 1+2+4+...+2(k-1) = 2k - 1 Step 3. Prove true for n = k+1, i.e. 1+2+4+...+2(k-1) + 2k = 2k+1 - 1 LHS = 2k - 1 + 2k (since...

Any laptop with the "Centrino" logo comes with a wireless adaptor. For about $1200 or so, you'd be looking for Centrino, 1.5Ghz Pentium-M, 256MB RAM (512 MB is much, much better if you can get it - if you can't I reccomend you upgrade as soon as you get the laptop, costs under $100 and it'll be...

not to mention America conductions their executions much more professionally than Singapore [ hanging? dont make me scoff how primitive] Sure. Keeping an inmate hanging for an average of 11 years on death row, with some waiting for over 20, is very, very professional. Making these people...

CP actually costs more than keeping a prisioner for life. That's only in America, where the law allows for endless appeals and delays in carrying out death sentences. The average time spent on the US death row is 11 years per inmate (2004 figure). 11 years of retrials, appeals, and time...

What was your class average? Unless the average was 95+, I wouldn't even think of dropping to General Maths. The test may have been relatively difficult, and besides - it's only the first one. I_F

Maths Extension 1 - 32/40 (rank 1). I_F

1. I'd suggest having one folder for each subject, with plastic sleeves for sheets and looseleaf paper, along with an exercise book. 2. Enjoy yourself along the way. Work hard, yes, but it's the penultimate year, so don't immerse yourself too deep in study. ;) I_F

Just because it's open book doesn't mean it's worth nothing. If you're flicking through a book, you're wasting time that could go to answering and checking questions. I_F

x squared = x + x + x.... x times deriving both sides 2x = 1 + 1 + 1... x times 2x = x 2 = 1 You can't divide through by a variable because the variable may be 0. In this case, 2 x 0 = 1 x 0. Taking the derivative f both sides is also inaccurate, since (x2 - x - 6) = (x - 6) when x = 0...

You shouldn't have to ask the same question in two different threads. For 1a, the answer is 336 ways, 8P3. I_F

Our first 4U assessment is tommorrow, and it's open book. I can't be bothered studying because there's only three people in the class and the other two are very likely to drop it. Besides, it's nearly holidays and I'm too burned out to care. I_F

1a) in how many ways can 3 horses win 1st, 2nd, and 3rd places in a race (trifecta) if there are 8 horses? I think this is a permutation - 8P3, or 8!/(8-3)! - 336 ways. I_F

I worked it out like this: If the question wants the probability of only one winning, then it is the probability of one winning the tournament plus the probability of the other winning the tournament, minus the chance that they will come equal first. That is, 1/4 + 1/3 - 1/12 = 1/2. If the...