Question

can anyone help me with mathematical induction please?

Answer 1

1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = quantity n times quantity six n squared minus three n minus one all divided by two

Answer 2

yes

Answer 3

can you put that in mathematical notation

Answer 4

okay so i did the left side and ended up with 18k^2-6k+5

Answer 5

yes of course

Answer 6

1^2+4^2+7^2+...+(n(6n^2-3n-1))/2

Answer 7

is there a restriction on n? like for all positive integers, or n>2 or some such? otherwise, what is it we are trying to prove by induction?

Answer 8

i need to do n=k n=k+1

Answer 9

i already did n=1

Answer 10

so you are trying to prove that this holds true for integers 1,2,3,... just so i can catch up, let n=1 to verify.

(3n-2)^2 = n(6n^2-3n-1)/2
(3-2)^2 = (6-3-1)/2
(1)^2 = (2)/2 , is good

now, let n=k is simple enough, right?

1^2+4^2+7^2+...+(3k-2)^2 = k(6k^2-3k-1)/2

we good to here?

Answer 11

yes and now you plug in k+1

Answer 12

not quite. we add the next term to each side: the next term is (3(k+1)-2)^2

Answer 13

alright

Answer 14

let P(k) = 1^2+4^2+7^2+...+(3k-2)^2

therefore, P(k) = n(6n^2-3n-1)/2

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

Answer 15

now we need to reduce it to a format that looks like we replaced k with k+1

Answer 16

pfft, i got some ns straggling in my post :/

Answer 17

(3k+3-2)^2

Answer 18

ns?

Answer 19

we need to show that:\[\frac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2} \]

Answer 20

yeah, when i was rewriting the left side in a few posts up i forgot to replace n with k :/

Answer 21

how did you get that

Answer 22

how did i get what?

Answer 23

ohh it's okay no worries ... the one above :yeah, when i was rewriting the left side in a few posts up i forgot to replace n with k :/

Answer 24

mathematical induction is all about form, if we know one form is good, then we need to develop a way to get the next term, the (k+1)th term, is in the same exact form.

Answer 25

as such, we know that the form:\[\frac{k(6k^2-3k-1)}{2}\]is of good form
we need to add the next term to it, the next term is defined as:\[(3(k+1)-2)^2\]
and show that the added term simplifies/reduces to the same form, namely:\[\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]
as if we simply replaced k by (k+1)

Answer 26

yea that was my problem i wasn't sure how to add it to the fraction, so basically you just put it on top of the 2

Answer 27

lets follow the setup this way, do we agree that:\[1^2+4^2+7^2+...+(3k-2)^2=\frac{k(6k^2-3k-1)}{2}\]
???

Answer 28

we know it is true for some k, namely k=1

Answer 29

yes

Answer 30

the next term to add, the (k+1)th term is simply added to each side:
\[1^2+4^2+7^2+...+(3k-2)^2 +\underbrace{(3(k+1)-2)^2}\\~~~~~~~~~~~=\frac{k(6k^2-3k-1)}{2}+\underbrace{(3(k+1)-2)^2}\]

Answer 31

okay i see what you mean

Answer 32

the left side is simply redundant, we need to show that the right side simplifies to the proper FORM

Answer 33

so we don't do anything to the left?

Answer 34

nope, the left is just listing all the terms up to the (k+1)th term. the right side is what we need to deal with.

Answer 35

okay :)

Answer 36

so, the rest is just algebraing it to death to get it transformed into:\[\frac{k(6k^2-3k-1)}{2}+\underbrace{(3(k+1)-2)^2}=^?\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]

Answer 37

yea that's the part i need help on if you don't mind.. I need to show the steps

Answer 38

well, get it all into one fraction to start with ... any ideas how to do that?

Answer 39

i was thinking of multiplying by 2/2

Answer 40

thats the way to go about it yes :) the process is really elementary, just cluttered is all

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

what would you attempt next?

Answer 41

distribute it so: 2(3k+2)-2)^2
2(3k)^2

Answer 42

3k*3k=2(9k)=18k

Answer 43

expanding it all out to get rid of the groupings would be a fine idea, then we can attempt to factor out a (k+1), lets ignor the /2 part since it has no part to play in the process at this time
\[k(6k^2-3k-1)+2(3(k+1)-2)^2\]
\[6k^3-3k^2-k+(12(k+1)-8)^2\]
\[6k^3-3k^2-k+(12k+12-8)^2\]
\[6k^3-3k^2-k+(12k+4)^2\]
\[6k^3-3k^2-k+144k^2+48k+16\]
\[6k^3+141k^2+47k+16\]
now to see if i did it right by checking with the wolf lol

Answer 44

how did you get 12 and 8

Answer 45

i tried to do something fancy and messed myself up is all :) if i had done it normally, we would have gotten to
\[6 k^3+15 k^2+11 k+2\]

Answer 46

that's what i got but over 2

Answer 47

but they don't match that's the problem

Answer 48

are they supposed to match?

Answer 49

yeah, the over 2 at the moment is clutter; what doesnt match?

Answer 50

my left and right... for my left i got 18k^2-6k+5 and for my right i got 6k^3+15k+11k+2/2

Answer 51

there is no 'left' to deal with.

lets factor out a (k+1), either by longhand, or synthetic division

Answer 52

so i wasn't supposed to solve the left

Answer 53

we are not on the same page at the moment, ive either forgotten what a left is, or im thinking of something else.

so far we have gotten:\[\frac{k(k^2-3k-1)}{2}+(3(k+1)-2)^2\]and worked it down to\[\frac{6k^3+15k+11k+2}{2}\]
since we need a (k+1) times [....]/2, lets factor out a (k+1) by some means and see what that leads to


6k^3+15k+11k+2
0 -6 -9 -2
-1 6 9 2

Answer 54

ohhh okay got it.... so it would be:

Answer 55

isnt 15k squared because in mine it's squared

Answer 56

yes :) typing and mathing do not always coexist in peaceful harmony

so, by synthetic division method

6k^3+15k^2+11k+2
0 -6 -9 -2
-1 6 9 2 0

this reworks to\[\frac{(k+1)(6k^2+9k+2)}{2}\]