Recent content by racket

I get it! Thank you, it's early and for some reason I didn't see it. Do you mind explaining one more thing? Why exactly does the inductive step make the left side what it's needed to be? Is that the same for all questions like this one?

Ah I'm sorry this is still confusing to me. Why exactly do you add (k+1)2^k+1 to both sides? On the left side do you change the i's to k's ?

Once I've done that, would I get: k(2^k+2) + 2 On the right side? How do I complete the proof from this point?

When I do that with the base case as 1 I get: 1 * 2^1 = (1 -1) * (2^n+1) + 2 2 = 2 Okay so, here's my work so far: I'm assuming n = m when m >= 1 So n + 1. And (m - 1) * (2^m+1) + 2 = m(2^m+2) + 2 Now I'm stuck but am I on the right track?

Yes that's exactly what I'm trying to prove, thanks for formatting it for me!

Hey! I've just started to learn some mathematical induction and it's proving quite tricky. Hopefully someone could help me out a little with my questions! I don't know how to format question on this forum but I'll try my best to accurately represent it: n $$sigma$$ i * 2^i = (n-1) * (2^n+1) +...