Show that:
\[1*3+2*3^{2}+...+n3^{n} =\frac{ (2n-1)3^{n+1}+1 }{ 4 }\]
for every number n where \[n \in \mathbb{N}, \mathbb{N}>0\]
If you try the induction formula with n=2,3 it get's wrong so is it the right conclution to just say that it isn't correct for all positive n ?

- anonymous

- katieb

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- anonymous

PMI will do it..)

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

- anonymous

So the left expression and the right doesn'r give the same value, doesn't that mean that the right expression is wrong for the left serie?

- myininaya

You just gave a counterexample @frx

- myininaya

You are right @frx
The expressions aren't equal.

- anonymous

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

- myininaya

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

- myininaya

I wonder if that works

- anonymous

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

- myininaya

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

- myininaya

I would do it both ways for your teacher.
Like you can do it the way the teacher put it and this way also.
It will be fun.
Counterexample for the +1 and a proof for the +3

- anonymous

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

- anonymous

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

- myininaya

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

- anonymous

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

- anonymous

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

- myininaya

Yep assume for some integer k>=1 that the equation holds.
Then you must show that your equation holds for k+1 :)

- anonymous

\[\sum_{a=1}^{k} a3^{a} = \frac{ (2k-1)3^{(k+1)}+3 }{4 }\]
\[\sum_{a=1}^{k+1} a3^{a} = \frac{ (2k+1)3^{(k+2)}+3 }{4 }\]

- myininaya

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

- anonymous

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

- myininaya

We get to assume:
\[1 \cdot 3 +2 \cdot 3^2+ \cdots + k \cdot 3^k=\frac{(2k-1)3^{k+1}+3}{4}\]
Now if we had (looking at the left side of your equation)
\[1 \cdot 3 +2 \cdot 3^2+ \cdots + k \cdot 3^k+(k+1) \cdot 3^{k+1}\]
By induction we can say this following thing in brackets is equal to?
\[[1 \cdot 3 +2 \cdot 3^2+ \cdots + k \cdot 3^k]+(k+1) \cdot 3^{k+1}\]

- myininaya

Remember what we get to assume about that thing in brackets (I wrote it as the first thing in that post just above ^)

- myininaya

\[\frac{(2k-1)3^{k+1}+3}{4}+(k+1) \cdot 3^{k+1}\]
Now all you have to do is do a little algebra to show this is equal to the right hand side of equation where you have:
\[\frac{(2[k+1]-1)3^{[k+1]+1}+1}{4}\]

- anonymous

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

- anonymous

Really, thank you so much!

- anonymous

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

- myininaya

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

- anonymous

Right, I've used another type of induction in analysis, but it's the way to deal with sum induction atleast

- myininaya

Yep :)

- myininaya

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

- anonymous

Actually i'm taking a course just called algebra and one analysis, but i'm in school i Sweden so the US conterpart i guess is called that :)

- myininaya

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

- anonymous

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

- myininaya

np/

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