## anonymous 3 years ago 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 ?

1. anonymous

PMI will do it..)

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. 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?

6. myininaya

You just gave a counterexample @frx

7. myininaya

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

8. anonymous

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

9. myininaya

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

10. myininaya

I wonder if that works

11. anonymous

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

12. myininaya

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

13. 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

14. anonymous

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

15. anonymous
16. myininaya

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

17. anonymous

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

18. anonymous

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

19. myininaya

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

20. anonymous

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

21. myininaya

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

22. anonymous

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

23. 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? $[1 \cdot 3 +2 \cdot 3^2+ \cdots + k \cdot 3^k]+(k+1) \cdot 3^{k+1}$

24. myininaya

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

$\frac{(2k-1)3^{k+1}+3}{4}+(k+1) \cdot 3^{k+1}$ 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. anonymous

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

27. anonymous

Really, thank you so much!

28. anonymous

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

29. myininaya

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

30. anonymous

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

31. myininaya

Yep :)

32. myininaya

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

33. 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 :)

34. myininaya

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

35. anonymous

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

36. myininaya

np/