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PMI will do it..)

p(1) = 3
p(k) =
\[\frac{ (2k-1)3^{k+1} +1}{ 4 }\]

We shuld Prove
p(k+1) =
\[\frac{ (2(k+1)-1)3^{k+1+1} +1}{ 4 }\]

What about p(2) it equals \[\frac{ 41 }{ 2 }\] but 2*3^2 = 18,

p(2)=18
n.3^n = 18

I think it is meant to be
\[\frac{(2n-1)3^{n+1}+3}{4}\]

I wonder if that works

You're right, now wonder it's didn't make any sense, it's +3 not +1

Yep it looks like it does :)
I got this to work for n=1,n=2 and n=3 :)

Well I don't know why but i don't get it right anyways. I get n=2 to be 21

http://www.wolframalpha.com/input/?i=%28%282n-1%293^%28n%2B1%29%2B3%29%2F4%2C+n%3D2

Right! And then the other side is 1*3^1+2*3^2=3+2(9)+3+18=21

Oh! of course, my bad didn't add the first 3 on the left side

So know I use induction to prove it, by setting n=k and n=k+1 right?

Ok. and you used the other side to get that also?

I'm not really familiar with the sum induction i don't really know how to prove it

Ohh! Now the coin fell down! I see now that it proves it for all positive integers, awesome

Really, thank you so much!

By the way, is this the way to treat all induction problems?

I think I will say yes. There is different kinds of induction though I think.

Yep :)

So my guess is that you are taking discrete math and cal 1 at the same time.

Ok. Well good luck with all your stuff. I hope you do well. :)

Thank you, and ones more thank you for your help :)

np/