Question

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 ?

Answer 1

PMI will do it..)

Answer 2

p(1) = 3

p(k) =

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

Answer 3

We shuld Prove

p(k+1) =

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

Answer 4

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

Answer 5

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?

Answer 6

You just gave a counterexample @frx

Answer 7

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

Answer 8

p(2)=18

n.3^n = 18

Answer 9

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

Answer 10

I wonder if that works

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

http://www.wolframalpha.com/input/?i=%28%282n-1%293^%28n%2B1%29%2B3%29%2F4%2C+n%3D2

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

Really, thank you so much!

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

Yep :)

Answer 32

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

Answer 33

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 :)

Answer 34

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

Answer 35

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

Answer 36

np/