Recent content by Hat1324

Yeah I know I only need one base case, but I couldn't do the work above without showing n=0,1,2 so I included them anyway

OK. I think I have it. So what I ended up doing was brute forcing a base case of ##0 \le n \le 2## and testing it for 0, 1, and 2 (which holds). Then I showed that for all ##k \ge 2## ##3^{k+1} = 3*3^k## ##2^{k+1} = 2*2^k## ##k+1 \le (1.5)k## This got me to derive ##3^{k+1} \ge (k+1)2^{k+1}...

Heh. I'm following as best I can I really don't follow. :P I can't see how ∃ 3^{k+1} ≤ (k+1)2^{k+1} is any easier to disprove than ∀ 3^{k+1} ≥ (k+1)2^{k+1} is to prove.

I have and I wish. But we're required to prove this directly using induction

Things are easier in that the question is asking about scalar speed. What does it mean for the derivative when speed hits a peak/trough?

I think I've explained my steps a little wrong. The first thing I did after the base case was assume 3^k≥k2^k Then I setup 3^{k+1}=3(3k)≥3(k2^k)≥... But I simply cannot find something to substitute ... that is actually true

Homework Statement The question asks me to prove inductively that 3n ≥ n2n for all n ≥ 0. Homework EquationsThe Attempt at a Solution I believe the base case is when n = 0, in which case this is true. However, I cannot for the life of me prove n = k+1 when n=k is true. I start with: 3^k ≥...