At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

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/