Recent content by FritoTaco

Just by looking at the two functions ##f(n)## and ##g(n)##, I can see that ##f(n)## is smaller than ##g(n)## because of ##n^2##. To prove this. I attempted the following and I'm not sure if I'm right. ##log(n+5) \leq c \cdot log(25n^2)## for all ##n \geq n_0## ##log(n+5) \leq c * log(25) +...

Awesome! Thanks for helping me!

I’m at school, so I can’t write latex. But I see, the (k+1) is just a precedent of (k+2)! So you can leave it as that. I think that’s what you’re saying?

Not exactly. I googled it and it mentioned (n+1)! = (n+1)·n!. but I'm not sure if that's what you do? https://en.wikipedia.org/wiki/Recursive_definition

So this is the same as (k+2)!-1?

(k+1)! factors out (k+1)![(k+1)+1]

is a = (k+1)! b = -1 c = (k+1) d = (k+1)!

does separating give us: (k+1)!=1+(k+1) I will get back tomorrow if I can.

Yeah, in my first post I did two steps in one without explaining that part. But, you mention to fill the dots part, that's where I'm stuck. I don't know where to go from there?

Sorry, I follow what you are writing but I don't understand why I'm not ultimately trying to get [(k+1)!-1]+(k+1)(k+1)! to equal (k+2)!-1. I was following my example in my book which you can see in a screenshot i attached. Here is my work: P(k): 1\cdot1!+2\cdot2!+..+k\cdot k! = (k+1)!-1...

So, what you're telling me is instead of the RHS being Q(k+1)+p_k it should be Q(k)+p_k or how in your original comment left it as (k+1)!-1?

It's (k+1+1)! because (just like I did on the left side) I replaced the k in (k+1)! to be (k+1+ 1)!. Because of P(k+1) I'm replacing k with k+1. I think that's what you're asking me about?

Homework Statement Prove that 1\cdot1!+2\cdot2!+...+n\cdot n! = (n+1)!-1 whenever n is a positive integer. Homework EquationsThe Attempt at a Solution I'm having trouble simplifying towards the end of the proof. Proof: Let P(n) be the statement 1\cdot1!+2\cdot2!+...+n\cdot n! = (n+1)!-1...

My teacher never used this method so I don't know if I will actually use it but maybe we will. Thanks!

Awesome! Thanks for helping me. There were a few other problems I had the same issue with and couldn't remember what else I was supposed to do, but now it all works.