Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

Answer 1

The teacher says I only completed part 1 of question two

Answer 2

@amistre64 @ganeshie8 @Abhisar

Answer 3

@Hero

Answer 4

since n=1 is false, then you have proved the statement to be false and there is no need to carry it into the other steps as i see it

Answer 5

Thats What I though t but she said that I only did the first step

Answer 6

is sthere anything else to do?

Answer 7

wait, you are focusing on question #2 on your paper right?

Answer 8

Yes this is what the teacher said "For problem #2, you only completed the first step"

Answer 9

you showed that when n=1 it is true

showing that it is only true for 1 integer does not prove that it is true for anything else

Answer 10

once we know it is true for some integer, then we can make the proof more general and show that if it is true for some k, then it is true for k+1 by converting the k+1 term setup into the same format for the k term setup

Answer 11

the second step is to assume that it is true from some n=k and say

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

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

now we need to prove that P(k+1) = (k+1)[6(k+1)^2-3(k+1)-1]/2

Answer 12

How do we do so? @amistre64

Answer 13

well, add (3(k+1)-2)^2 to each side and then manipulate it with algebra into the FORM that we need to show.

Answer 14

Im a bit confused on what two sides

Answer 15

\[P(k)=\frac12k(6k^2-3k-1)\]
\[P(k)+\underbrace{(3(k+1)-2)^2}=\frac12k(6k^2-3k-1)+\underbrace{(3(k+1)-2)^2}\]
now show that the right side can be manilpulated algebraically into the same FORM as the kth term, namely:
\[P(k+1)=\frac12\color{red}{(k+1)}(6\color{red}{(k+1)}^2-3\color{red}{(k+1)}-1)\]

Answer 16

the equal sign has 2 sides, a left and a right

Answer 17

so would it be p(k+1)=(.5k+.5)(6k^2+36)-3k-3-1?

Answer 18

lets step thru it
\[given:\\\underbrace{1^2 + 4^2 + 7^2 + ...+(3k-2)^2}_{\large P(k)}=\frac12k(6k^2-3k-1)\]
\[add,~(3(k+1)-2)^2~to~each~side:
\\\underbrace{P(k)+(3(k+1)-2)^2}_{\large P(k+1)}=\frac12k(6k^2-3k-1)+(3(k+1)-2)^2\]
now, the left side of this is finished, we are done with it.

we simply need to get the right side of the equal sign to get into the same exact FORM as it was before so that it simply looks as if we replaced the original k stuff with k+1

Answer 19

the only thing we know is true, is the form itself

Answer 20

does:\[\frac12k(6k^2-3k-1)+(3(k+1)-2)^2=\frac12(k+1)(6(k+1)^2-3(k+1)-2)\]???

Answer 21

lets expand it out and try to get it to match up

Answer 22

\[\frac12k(6k^2-3k-1)+(3(k+1)-2)^2\]
\[\frac12k(6k^2-3k-1)+(3k+3-2)^2\]
\[\frac12k(6k^2-3k-1)+(3k+1)^2\]
\[\frac12(6k^3-3k^2-k)+(9k^2+6k+1)\]
\[\frac12(6k^3-3k^2-k+18k^2+12k+2)\]
\[\frac12(6k^3+15k^2+11k+2)\]
might as well factor out a k+1 and see what we end up with now
\[\frac12(k+1)(6k^2+9k+2)\]
now we need some guidance perhaps to see how to proceed

Answer 23

UmM

Answer 24

Would I not distribute the 1/2

Answer 25

now, we need the 1/2 as it is since that is the original form that we have it all in. we CANNOT change the form.

what we want is
\[6(k+1)^2-3(k+1)-2\]
\[6(k^2+2k+1)-3k-3-2\]
\[6k^2+12k+6-3k-3-2\]
this tells us how to work what we have into what we want now

Answer 26

Wel then when does it end

Answer 27

when we get it into the same form that we started with, only difference is that the original k needs to be a k+1

Answer 28

Cant we just ad +1 next to each k?

Answer 29

i just noticed that i was carrying a -2 instead of a -1, corrected as

we started with \[\frac12k(6k^2-3k-1)\]
we need to show that by adding in the next term, the k+1th term, we can conform the new addition into the same form, such that the k become k+1:\[\frac12(k+1)(6(k+1)^2-3(k+1)-1)\]

Answer 30

no, we cant just add a +1 to the ks and say that we have proved it./

Answer 31

we have to show, mathically, that the process can be tranformed into it

Answer 32

if we dont work the mathing, we havent proved anything and we are simply telling people to take our word for it

Answer 33

So what is the whole procesS in adition to what I wrote , I got confused

Answer 34

I dont know what you mean when you said you were carying a two

Answer 35

our starting format is
\[6(k+1)^2-3(k+1)-1\]
when i was mathing the right side i got to
\[6k^2+9k+2\]
then that ending 2 i carried over by mistake into saying that we wanted to show
\[6(k+1)^2-3(k+1)-2\]instead of the -1 end

Answer 36

i worked the mathing from our right side of:
\[\frac12k(6k^2-3k-1)+(3(k+1)-2)^2\]
into a more workable form of:
\[\frac12(k+1)(6k^2+9k+2)\]

Answer 37

all thats left is to show that we can get from:
\[6k^2+9k+2\implies6(k+1)^2-3(k+1)-1\]

Answer 38

why do we want to math it into that form?

Answer 39

because we know that form is good, and if it simply looks like we replaced the original k with k+1 ... we have shown that the system remains consistent.

Answer 40

we cant simply say, replace it ... we have to show that the math process logically transforms the added term into it

Answer 41

the basis step shows that it is true for one specific value.

the induction step shows that if it is true for some value, that we can demonstrate that it is true for the next value.

if its true for k, then we show that its true for k+1

if its true for k=1, then its true for k=2
if its true for k=2, then its true for k=3
if its true for k=3, then its true for k=4
etc, etc, etc

and all this is demonstrated in showing that the 'formula' remains consistent for k as it does for k+1

Answer 42

well, consistent for k+1 as it does for k :)