Recent content by ngluth

Awesome ! Thank you so much. I did understand the general concept. Your explanation has help clear up a few places that were a little fuzzy. I could complete an induction when the equation was equal but the inequalities are confusing. Thanks again. I appreciate the service you are providing...

why is the answer not 1+2(k+1) < 3^k+1 I have turned in this assignment so could you Please explain exactly how I would finish this problem. I have not yet completely understood how to prove that if it's true for k, then it's true for k+1. The problem before the one I got it answered. I...

Another question: P(n) 1^2 + 2^2 + 3^2 +…+ n^2 = n(n+1)(2n+1) / 6 Consider: P(1) n^1 = 1(n+1)[(2(1) +1] /6 1^1 = 1(2)(2+1) /6 1 = 6 / 6 or 1 Since 1 = 1 I have established P(1) P(K) 1^2 + 2^2 + 3^2 +…+ K^2 = K(K+1)(2K+1) /6 Implies P(k+1) 1^2 + 2^2 + 3^2 +…+ K^2 +...

That is what I figured out. Thanks What a help

Need some help with this induction problem. I have it thus far P(n) n> or equal ?, 1+2n < 3^n consider P(1) 1+2(1) < 3^1 = 4<3 no consider P(2) 1+2(2) < 3^2 = 5<9 yes since 5<9, P(2) is established Show that K > or equal 2 if P(k): 1+2k < 3 then P(k+1): 1+2(k+1) < 3^k+1...

I'm sending this again. Don't think it went the first time. Thanks for all your time and patience. I can now finish the assignment. This was the hardest problem. I just could not understand what was going on with it. Thanks. This is an awesome service you are providing. Thanks

Thank you for all your time and patience. Now I can finish the rest of the assignment. This was the hardest one and I just could not understand what they were asking. Thanks

k! = k(k-1)...2 * 1 I am really searching here I have been out of school for 35 years. I don't think they taught this stuff back then.

k! would be k * k+1 * k+2 * k+3 etc. and (k+1)! would be (K+1) * (k+2) * (k+3) etc

Left side means 2^k * 2^1 Got it Sorry In the last part are you asking me how to get (k+1)! and k! to equal each other?

OKay Now I see it. Now you asked How I get (k+1)! form k! I have to admit I have no idea.

How does it become 2*2^k

I don't understand how 2^(k+1) is = to 2*2^k

I am trying to understand induction and not having much luck. Here is my problem as I understand it Problem: n > or equal to 4, 2^n < n! Step 1) Prove it works for n=1 no, n=2 no, n=3 no, n=4 yes 16 < 24 step 2) assume it works for n=k 2^k < k! Step 3) prove it works for n= (k+1)...