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anonymous
 3 years ago
Show that:
\[1*3+2*3^{2}+...+n3^{n} =\frac{ (2n1)3^{n+1}+1 }{ 4 }\]
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
 3 years ago
Show that: \[1*3+2*3^{2}+...+n3^{n} =\frac{ (2n1)3^{n+1}+1 }{ 4 }\] 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 ?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0p(1) = 3 p(k) = \[\frac{ (2k1)3^{k+1} +1}{ 4 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We shuld Prove p(k+1) = \[\frac{ (2(k+1)1)3^{k+1+1} +1}{ 4 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What about p(2) it equals \[\frac{ 41 }{ 2 }\] but 2*3^2 = 18,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.2You just gave a counterexample @frx

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2You are right @frx The expressions aren't equal.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2I think it is meant to be \[\frac{(2n1)3^{n+1}+3}{4}\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2I wonder if that works

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You're right, now wonder it's didn't make any sense, it's +3 not +1

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2Yep it looks like it does :) I got this to work for n=1,n=2 and n=3 :)

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2I 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
 3 years ago
Best ResponseYou've already chosen the best response.0Well I don't know why but i don't get it right anyways. I get n=2 to be 21

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%28%282n1%293^%28n%2B1%29%2B3%29%2F4%2C+n%3D2

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2Right! And then the other side is 1*3^1+2*3^2=3+2(9)+3+18=21

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh! of course, my bad didn't add the first 3 on the left side

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So know I use induction to prove it, by setting n=k and n=k+1 right?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2Yep assume for some integer k>=1 that the equation holds. Then you must show that your equation holds for k+1 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{a=1}^{k} a3^{a} = \frac{ (2k1)3^{(k+1)}+3 }{4 }\] \[\sum_{a=1}^{k+1} a3^{a} = \frac{ (2k+1)3^{(k+2)}+3 }{4 }\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2Ok. and you used the other side to get that also?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not really familiar with the sum induction i don't really know how to prove it

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2We get to assume: \[1 \cdot 3 +2 \cdot 3^2+ \cdots + k \cdot 3^k=\frac{(2k1)3^{k+1}+3}{4}\] 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
 3 years ago
Best ResponseYou've already chosen the best response.2Remember what we get to assume about that thing in brackets (I wrote it as the first thing in that post just above ^)

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2\[\frac{(2k1)3^{k+1}+3}{4}+(k+1) \cdot 3^{k+1}\] 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
 3 years ago
Best ResponseYou've already chosen the best response.0Ohh! Now the coin fell down! I see now that it proves it for all positive integers, awesome

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Really, thank you so much!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By the way, is this the way to treat all induction problems?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2I think I will say yes. There is different kinds of induction though I think.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right, I've used another type of induction in analysis, but it's the way to deal with sum induction atleast

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2So my guess is that you are taking discrete math and cal 1 at the same time.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually 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
 3 years ago
Best ResponseYou've already chosen the best response.2Ok. Well good luck with all your stuff. I hope you do well. :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you, and ones more thank you for your help :)
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