Question

help please?

Answer 1

Use mathematical induction to prove the statement is true for all positive integers n.

The integer n3 + 2n is divisible by 3 for every positive integer n

Answer 2

\(n^3+2n\) ?

Answer 3

yes

Answer 4

and you have to use induction right? not some other method

Answer 5

yes induction

Answer 6

okay the first step is to show it is true if \(n=1\)
did you do that?

Answer 7

actually before we begin, do you know how to do a proof by induction?

Answer 8

Yes I just was not sure for this one what I was supposed to be doing because it says they have to be divisible by 3.

Answer 9

first you do the case \(n=1\)
that is, say that \(1^3+2\times 1\) is divisible by \(3\), which it is, because it is 3

Answer 10

Oh okay and you do that for each term? like 2, 3, 4, and so on?

Answer 11

oh no

Answer 12

the proof is "by induction"

Answer 13

you assume it is true if \(n=k\) and then prove under that assumption, that it is true if \(n=k+1\)

Answer 14

write down what you get to assume if \(n=k\) which really means just replace \(n\) by \(k\) and say "assume \(k^3+2k\) is divisible by 3

Answer 15

then replace \(k\) by \(k+1\) and try to prove that
\[(k+1)^2+2(k+1)\] is divisible by 3

Answer 16

that is going to take a little algebra
let me know if i have lost you yet

Answer 17

I have got \[k^3+2k\] and then I made it \[(k+1)^3+2(k+1)\], but how do you prove that it is divisible by three? I always have trouble solving the k+1 form.

Answer 18

expand, then see if you can find a \(k^3+2k\) in the expansion

Answer 19

that is the algebra part

Answer 20

\[(k+1)^3+2(k+1)=...=k^3+3 k^2+5 k+3\]

Answer 21

So should that be like \[(k+1)(k+1)(k+1)+2k+2\] and do foil on the \[(k+1)^3\]

Answer 22

the \(...\) part is the algebra

Answer 23

lol foil

Answer 24

haha ik that is how i remember it

Answer 25

\(a+b)^3=a^3+3a^2b+3ab^2+b^3\) you get
\[k^3+3k^3+3k+1\]

Answer 26

okay ad then add 2k+2

Answer 27

right

Answer 28

let me know when you get \(k^3+2k^2+5k+3\)

Answer 29

now comes the part where you find the \(k^3+2k\) in there

Answer 30

wait it should be 3k^2

Answer 31

yeah what you said, my algebra is bad

Answer 32

\[k^3+3 k^2+5 k+3\]

Answer 33

yes and then what do oyu do to find the k^3+2k?

Answer 34

now write is as \(k^3+2k+\text{whatever is left}\)

Answer 35

\(k^3+2k+3k^3+3k+3\) is what you should get

Answer 36

\[k^3+2k-3k^2+3k+3?\]

Answer 37

yes, but all plus signs

Answer 38

oh so you change the -3 to a positive

Answer 39

it was never negative

Answer 40

whoops wrote it down wrong

Answer 41

now you are almost done
you get to assume that \(k^3+2k\) is divisible by 3 "by induction"
the rest is clearly divisible by 3

Answer 42

\[k^3+2k+3k^2+3k+3=k^3+2k+3(k^2+k+1)\]

Answer 43

\[\overbrace{k^3+2k}^{\text{by induction}}+\overbrace{3(k^2+k+1)}^{\text{ obvious}}\]

Answer 44

since both parts are divisible by 3, the whole thing is
okay?

Answer 45

wait so that is the answer?

Answer 46

yes, you are done at that step

Answer 47

Oh okay thank you so much.

Answer 48

each part is divisible by 3, the first "by induction" and the second because there is a common factor of 3, so the whole thing is
yw