Question

Help with mathematical induction?

Answer 1

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.
\[1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{ n(6n^2 -3n-1) }{2 }\]

Answer 2

@ParthKohli ?

Answer 3

I know how to do this, but I'm really unwell right now. @iambatman Care to help out a fellow batman? :)

Answer 4

Help.. :/

Answer 5

DO I need to do n=x+1 after I do n=1? Or do I do the assume thing?

Answer 6

@RadEn ?????

Answer 7

@jdoe0001 @Light&Happiness
Somebody, please.. ._.

Answer 8

the induction base is: for n=1
left side:
\[\large 1^2 =1 \]
right side:
\[\large \frac{1(6\cdot1^2-3\cdot1-1)}{2}=\frac{6-4}{2}=1. \]
so the induction base case is true.

Answer 9

assuming the original statement is true (inductive hypothesis) then we have to prove (inductive thesis):
\[\large 1^2+4^2+\dots+(3n-2)^2+(3n+1)^2=
\frac{(n+1)[6(n+1)^2-3(n+1)-1]}{2} \]

Answer 10

\[\large =\frac{(n+1)(6n^2+9n+2)}{2} \]

Answer 11

can u dot it?

Answer 12

Can I what? :p

Answer 13

prove the inductive thesis.

Answer 14

No... v_v

Answer 15

How? Does it have something to do with the (3n-2)^2 + (3n+1)^2?

Answer 16

Ugh, he left. @jim_thompson5910 , or @Hero ?

Answer 17

@dumbcow , @.Sam.

Answer 18

It looks like you've already done the case n = 1. So I'll move on



Assume the equation holds true when n = k, ie


\[1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = \frac{ k(6k^2 -3k-1) }{2 }\]


The goal is to show that the equation also holds true when n = k+1 (based on the assumption that the equation is true when n = k)

Answer 19

I filled in the n's with n-1, but I don't understand how to finish that step, so I haven't.. @jim_thompson5910

Answer 20

Plug in n = k+1 to get

\[1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{ n(6n^2 -3n-1) }{2 }\]

\[1^2 + 4^2 + 7^2 + ... + (3(k+1) - 2)^2 = \frac{ (k+1)(6(k+1)^2 -3(k+1)-1) }{2 }\]

\[1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 + (3(k+1) - 2)^2 = \frac{ (k+1)(6(k+1)^2 -3(k+1)-1) }{2 }\]

\[\sum\left(\text{Terms up to n = k}\right) + (3(k+1) - 2)^2 = \frac{ (k+1)(6(k+1)^2 -3(k+1)-1) }{2 }\]

\[\frac{ k(6k^2 -3k-1) }{2 } + (3(k+1) - 2)^2 = \frac{ (k+1)(6(k+1)^2 -3(k+1)-1) }{2 }\]

see where this is going?

Answer 21

Not really...

Answer 22

ok one sec and I'll write out the rest

Answer 23

So you are messing with the left side?

Answer 24

correct, I'm trying to turn the left side into the right side

Answer 25

So you add the original problem to (3k+1)^2?

Answer 26

I'm getting stuck though, so I might try manipulating the right side to turn into the left

Answer 27

ok I think I figured it out, gotta write it up though

Answer 28

I can't get them to equal up to eachother.. I'm so lost.

Answer 29

ok, it's quite long and I couldn't get it to fit on here so I made a pdf of the entire process (of just the inductive step, not the n = 1 case, but that's trivial really)

Answer 30

see attached

Answer 31

it's definitely a lot, so let me know if you have any questions on it

Answer 32

What did you do when /you went from:\[(k+1)(6k^2 +12k+6-3k-3-1) \] to:\[(k+1) ((6k^2-3k-1)+12k+6-3)\]?

Answer 33

Why did it get factored like that?

Answer 34

when you assume the equation is true for n = k, you get

\[1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = \frac{ k(6k^2 -3k-1) }{2 }\]

Answer 35

Notice how there's the term \[\Large 6k^2 - 3k - 1\]

I pulled that out because I wanted to have k times that over 2

Answer 36

that way, I would then have the sum from n = 1 to n = k (which is why this all depends on the assumption)

Answer 37

Here's the template