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

1. Yahoo! Group Title

PMI will do it..)

2. Yahoo! Group Title

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

3. Yahoo! Group Title

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

4. frx Group Title

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

5. frx Group Title

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?

6. myininaya Group Title

You just gave a counterexample @frx

7. myininaya Group Title

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

8. Yahoo! Group Title

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

9. myininaya Group Title

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

10. myininaya Group Title

I wonder if that works

11. frx Group Title

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

12. myininaya Group Title

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

13. myininaya Group Title

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

14. frx Group Title

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

15. frx Group Title
16. myininaya Group Title

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

17. frx Group Title

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

18. frx Group Title

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

19. myininaya Group Title

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

20. frx Group Title

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

21. myininaya Group Title

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

22. frx Group Title

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

23. myininaya Group Title

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? $[1 \cdot 3 +2 \cdot 3^2+ \cdots + k \cdot 3^k]+(k+1) \cdot 3^{k+1}$

24. myininaya Group Title

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

25. myininaya Group Title

$\frac{(2k-1)3^{k+1}+3}{4}+(k+1) \cdot 3^{k+1}$ 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: $\frac{(2[k+1]-1)3^{[k+1]+1}+1}{4}$

26. frx Group Title

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

27. frx Group Title

Really, thank you so much!

28. frx Group Title

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

29. myininaya Group Title

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

30. frx Group Title

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

31. myininaya Group Title

Yep :)

32. myininaya Group Title

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

33. frx Group Title

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 :)

34. myininaya Group Title

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

35. frx Group Title

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

36. myininaya Group Title

np/