Question

Write a proof by mathmatical induction of the statement: 1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3

Answer 1

anyone know?

Answer 2

your stuff is something, man!! no answer yet!!

Answer 3

what have you tried

Answer 4

i keep getting stuck. I really have no idea if im starting right or what.

Answer 5

this is a fairly standard problem. what is the first thing you did?

Answer 6

you get 10=10 with n=2

Answer 7

how" (2n-1)^2 = (2.2 -1)^2 = 3^2 =9

Answer 8

\[1^2+3^2=1+9=10\]

Answer 9

the LHS is a sum

Answer 10

are you waiting for someone to do the problem for you?

Answer 11

I would like to know how to go about solving the problem, or at least get a good link to an online guide.

Answer 12

first do the basis step

Answer 13

I have a solution and it has all steps. I just don't know how to let the student fill in some of the blanks. I'd have to give the whole thing.

Answer 14

i can work backwards and learn that way. at this point im so confused, i need to know what the correct method is.

Answer 15

https://en.wikipedia.org/wiki/Mathematical_induction

Answer 16

your post vanished?

Answer 17

thanks for posting. got it

Answer 18

uw!

Answer 19

thanks, im working through it now. ill let you know if i have any questions.

Answer 20

What's important here is the methodology on how to go about doing the mathematical induction. Pay attention to the "case of n=1" logic and the assumption of that it works for n=k. The reason that induction works is that k is 1 in the beginning, and then it works for "2". If it works for "2", it works for "3", and so on.

Answer 21

If you use this general logic and the setting up of an induction proof and use this as a guide, you have a good chance of doing an induction problem on your own now.

Answer 22

You show that it works for the case of n=1 and then you assume it works for n=k. Then you show that if you assume n=k and you can make it work for n=k+1, you've made your proof work.

So, for n=1:

[2(1)-1]^2 = 1 and (1(2(1)-1)(2(1)+1))/3 = 1

So it works for n=1. Now assume it works for n=k. Add [2(k+1)-1]^2 to both sides.

For the right side:

(k(2k-1)(2k+1))/3 + [2(k+1)-1]^2 =

(k(2k-1)(2k+1))/3 + 3([2(k+1)-1]^2)/3 =

(4k^3 - k)/3 + 3[(2k + 1)^2]/3 =

(4k^3 - k)/3 + (12k^2 + 12k + 3)/3 =

(4k^3 + 12k^2 + 11k + 3)/3 =

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

((k + 1)[2(k + 1) - 1][2(k + 1) + 1])/3

But this is just what we already assumed but now for the case n = k + 1, so we have now completed our proof.

Answer 23

what about the left side?

Answer 24

You added:

[2(k+1)-1]^2

to both sides. You don't have to do anything to the left side. It's just the continuation of the "k" case but now for k + 1.

Answer 25

It's just the next term in the sequence. It's the right side you prove with. The left side just shows the continuation.

Answer 26

btw, Good luck to you in all of your studies and thx for the recognition! @comput313

Answer 27

so the right side is 8k^2+2?

Answer 28

You see, once you make the assumption that it works for n=k, you can use laws of math to add the same thing to both sides and thereby keep the equality. You just re-write the right side into a format, with steps, that show it's the n= k + 1 case.

Answer 29

that comes out as 34=14??

Answer 30

Is it starting to make sense now, @comput313 ? Or are you going through my steps at this time?

Answer 31

i substituted in k=2 to test, and came out with 14=34.

Answer 32

I substituted in k=2 and came up with 10 = 10. I just logged back on because Openstudy kept dropping my connection. Let's try to stay in touch on this problem until you are comfortable with it.

Answer 33

For the left side i have (2k-1)^2+(2(k+1)-1)^2.