anonymous
  • anonymous
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 ?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
PMI will do it..)
anonymous
  • anonymous
p(1) = 3 p(k) = \[\frac{ (2k-1)3^{k+1} +1}{ 4 }\]
anonymous
  • anonymous
We shuld Prove p(k+1) = \[\frac{ (2(k+1)-1)3^{k+1+1} +1}{ 4 }\]

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anonymous
  • anonymous
What about p(2) it equals \[\frac{ 41 }{ 2 }\] but 2*3^2 = 18,
anonymous
  • 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
  • myininaya
You just gave a counterexample @frx
myininaya
  • myininaya
You are right @frx The expressions aren't equal.
anonymous
  • anonymous
p(2)=18 n.3^n = 18
myininaya
  • myininaya
I think it is meant to be \[\frac{(2n-1)3^{n+1}+3}{4}\]
myininaya
  • myininaya
I wonder if that works
anonymous
  • anonymous
You're right, now wonder it's didn't make any sense, it's +3 not +1
myininaya
  • myininaya
Yep it looks like it does :) I got this to work for n=1,n=2 and n=3 :)
myininaya
  • 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
  • anonymous
Well I don't know why but i don't get it right anyways. I get n=2 to be 21
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=%28%282n-1%293^%28n%2B1%29%2B3%29%2F4%2C+n%3D2
myininaya
  • myininaya
Right! And then the other side is 1*3^1+2*3^2=3+2(9)+3+18=21
anonymous
  • anonymous
Oh! of course, my bad didn't add the first 3 on the left side
anonymous
  • anonymous
So know I use induction to prove it, by setting n=k and n=k+1 right?
myininaya
  • myininaya
Yep assume for some integer k>=1 that the equation holds. Then you must show that your equation holds for k+1 :)
anonymous
  • 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
  • myininaya
Ok. and you used the other side to get that also?
anonymous
  • anonymous
I'm not really familiar with the sum induction i don't really know how to prove it
myininaya
  • 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
  • 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
  • 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
  • anonymous
Ohh! Now the coin fell down! I see now that it proves it for all positive integers, awesome
anonymous
  • anonymous
Really, thank you so much!
anonymous
  • anonymous
By the way, is this the way to treat all induction problems?
myininaya
  • myininaya
I think I will say yes. There is different kinds of induction though I think.
anonymous
  • anonymous
Right, I've used another type of induction in analysis, but it's the way to deal with sum induction atleast
myininaya
  • myininaya
Yep :)
myininaya
  • myininaya
So my guess is that you are taking discrete math and cal 1 at the same time.
anonymous
  • 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
  • myininaya
Ok. Well good luck with all your stuff. I hope you do well. :)
anonymous
  • anonymous
Thank you, and ones more thank you for your help :)
myininaya
  • myininaya
np/

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