1. ## Induction

Okay I must prove

Base n=1, K=1
1+4+7......+(3n-2) = n(2n-1)

(3k-2) = k(2k-1)
(3k-2)+(k+1) = (k+1)(2k+1-1)
= (k+1)(2k+2-1)
= (k+1)(2k+1)

now I am lost, please show me what to do!

2. Hello, CPR!

Where are you getting these problems?

Most of them are simply not true . . .

$\displaystyle 1+4+7 + \hdots + (3n-2) \:= \:n(2n-1)$

Did you try the formula at all?

For $\displaystyle n = 5$, the left side is: .$\displaystyle 1 + 4 + 7 + 10 + 13 \;=\;35$

The right side is: .$\displaystyle 5(2\!\cdot\!5-1) \;=\;45$

So they are lying to us . . .

3. We had to make up problems then solve using mathematical induction. I guess I wasnt doing a good job making them up.

This is one I have that should work

1+5+9.......+(4n-3)= n(3n-2)

Okay,
Base is n=1, k+1=1

1+5+9.....+(4n-3)+(k+1)= (k+1)(3k+1-2)
= (K+1)(3k-1)

Okay, please show me what to do now,for I'm lost I know I must make LHS equal RHS, how do I do this? Thanks

4. Well, CPR, it's not that hard to make problems like that, if you use the formula of arithmetic progress (which Gauss invented at the age of 10!):
$\displaystyle a_{1}+a_{2}+...+a_{n}=$ $\displaystyle \frac{a_{1}+a_{n}}{2}n$
Now, you can set up the LAST factor of your sum, let's say (4n-3), which is at your problem. For n=1, it's 4x1-3=1, for n=2 it's 4x2-3=5 and so on, therefore your sum indeed is 1+5+9+...+(4n-3). Now, the right side of your problem will be found by using the Gauss' formula. Hence must be $\displaystyle \frac{1+(4n-3)}{2}n=(2n-1)n$.

So, now, you know your -correct- right side of your problem, and you can apply induction.

Take a look at some similar problems:
1. 1+7+13+...+(6n-5) = n(3n-2)
2. 5+7+9+...+(2n+3)= n(n+2)
3. 4017+6025+...+(2008n+2009)=n(1004n+3013)

PS I hope i didn't make any mistake on the operations :-).

5. Thanks, but I still do not know what to do. Why do I need to divide by 2? Why has n(3n-2) become (2n-1)? As you can see I'm totally confused on your explanation.

6. Hello, CPR!

You must come up with a statement that is true.

$\displaystyle 1 + 5 + 9 + \hdots + (4n-3) \:= \:n(3n-2)$ . . . . not true!

Do you ever check your formulas?
This one isn't true either . . .

For $\displaystyle n = 4$, the left side is: .$\displaystyle 1 + 5 + 9 + 13 \:=\:28$

But the right side is: .$\displaystyle 4(3\!\cdot\!4-2) \;=\;40$

Here's the correct formula:

$\displaystyle 1 + 5 + 9 + \hdots + (4n-3) \;=\;n(2n-1)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

And you evidently don't understand the induction process.

When we write the $\displaystyle S(k+1)$ statement,
. . it does not mean "stick in a plus-one somewhere."

It means: Replace $\displaystyle k$ with $\displaystyle k+1$ . . . there's a big difference.

7. As you were shown, the series is $\displaystyle 1+5+9+....+(4n-3)=n(2n-1)$

Check for n=1:

1(2(1)-1)=1..........true.

Assume true for $\displaystyle P_{k}$. Therefore,

$\displaystyle 1+5+9+........+(4k-3)=k(2k-1)$

We must show it is true for $\displaystyle P_{k+1}$

That is, we have to show that $\displaystyle 1+5+9+...+(4(k+1)-3)=(k+1)(2(k+1)-1)$

Let's add 4k+1 to both sides and see what happens:

$\displaystyle 1+5+9+......+(4k-3)+(4k+1)=k(2k-1)+(4k+1)$

The right side: $\displaystyle k(2k-1)+(4k+1)=2k^{2}+3k+1=(k+1)(2k+1)=\boxed{(k+1)(2(k +1)-1)}$

This shows that $\displaystyle P_{k+1}$ is true and the proof is complete.

See how the k in our previous series is replaced with a k+1?. I stepped through it hoping you can see it and can do another. You can't just throw it in. It has to be shown.

8. Thank you so very much Galactus! For taking the time to explain it and not be critical. I am sorry that the formulas were not correct. But you corrected it and showed me how to do it. That is what I needed. I now see it and understand it and the process. Again thanks! I got it!