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frx
Group Title
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 ?
 one year ago
 one year ago
frx Group Title
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 ?
 one year ago
 one year ago

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Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
PMI will do it..)
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
p(1) = 3 p(k) = \[\frac{ (2k1)3^{k+1} +1}{ 4 }\]
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
We shuld Prove p(k+1) = \[\frac{ (2(k+1)1)3^{k+1+1} +1}{ 4 }\]
 one year ago

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

frx Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
You just gave a counterexample @frx
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
You are right @frx The expressions aren't equal.
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
p(2)=18 n.3^n = 18
 one year ago

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

myininaya Group TitleBest ResponseYou've already chosen the best response.2
I wonder if that works
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
You're right, now wonder it's didn't make any sense, it's +3 not +1
 one year ago

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

myininaya Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Well I don't know why but i don't get it right anyways. I get n=2 to be 21
 one year ago

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

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

frx Group TitleBest ResponseYou've already chosen the best response.0
Oh! of course, my bad didn't add the first 3 on the left side
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
So know I use induction to prove it, by setting n=k and n=k+1 right?
 one year ago

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

frx Group TitleBest ResponseYou've already chosen the best response.0
\[\sum_{a=1}^{k} a3^{a} = \frac{ (2k1)3^{(k+1)}+3 }{4 }\] \[\sum_{a=1}^{k+1} a3^{a} = \frac{ (2k+1)3^{(k+2)}+3 }{4 }\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Ok. and you used the other side to get that also?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
I'm not really familiar with the sum induction i don't really know how to prove it
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
We 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? \[[1 \cdot 3 +2 \cdot 3^2+ \cdots + k \cdot 3^k]+(k+1) \cdot 3^{k+1}\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Remember what we get to assume about that thing in brackets (I wrote it as the first thing in that post just above ^)
 one year ago

myininaya Group TitleBest 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: \[\frac{(2[k+1]1)3^{[k+1]+1}+1}{4}\]
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Ohh! Now the coin fell down! I see now that it proves it for all positive integers, awesome
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Really, thank you so much!
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
By the way, is this the way to treat all induction problems?
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
I think I will say yes. There is different kinds of induction though I think.
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Right, I've used another type of induction in analysis, but it's the way to deal with sum induction atleast
 one year ago

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

frx Group TitleBest ResponseYou've already chosen the best response.0
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 :)
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Ok. Well good luck with all your stuff. I hope you do well. :)
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Thank you, and ones more thank you for your help :)
 one year ago
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