Question

Help? :c
Use mathematical induction to prove the statement is true for all positive integers n.

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

Answer 1

@phi

Answer 2

plug in (x+1) into the formula n^3 + 2n
what do you get ?

Answer 3

\[x^3+3x^2+6x+4\]

Answer 4

that does not look right. can you show your steps ?

Answer 5

My bad. \[x^3+3x^3+5x+3\]

Answer 6

ok, now with that accomplished, we can do the proof by induction.

1) show that \[ n^3 +2n \text{ is evenly divisible by 3}\] is true for n=1

Answer 7

1^3+2(1)=1+2=3

Answer 8

and 3 is evenly divisible by 3

next step
2) assume
\[ n^3 +2n \text{ is evenly divisible by 3} \]
for all numbers x ≤ n

Answer 9

we use step 2 this way:
we let x= n+1, and use that value in the formula. As you figured out, we get
\[ n^3+3n^3+5n+3 \]
(sorry about switching the n's and x's)
we can rewrite your result using 5n= 3n + 2n
\[ n^3+2n + 3n^3+3n+3 \]

Answer 10

notice that the first 2 terms
\[ n^3 +2n\]
is assumed to be divisible by 3. (In other words, according to step 2, we assume the formula works for all numbers ≤ n.)

also, we see the last 3 terms
\[ 3n^3 +3n+3= 3(n^3+n+1) \]
is evenly divisible by 3

therefore the formula worked for x= n+1

Answer 11

in summary
1) formula is divisible by 3 for n=1
2) formula is divisible by 3 for n= n+1
therefore the formula is true for all n≥1

Answer 12

How did we prove that n^3+2n was?

Answer 13

We don't prove that n^3+2n is evenly divisible by 3. We ASSUME it is true for some n

what we prove is that IF n^3+2n is divisible by 3, then it is also true for x= n+1
(which we did)

the idea is that if n^3+2n is divisible by 3, and it is also true for n= n+1 (no matter what n is), then the formula works for n+2, n+3, n+4, and so on....

Answer 14

by step 1, we know the formula works for n=1

by step 2, if the formula works for n=1, it will work for n=1+1 = 2

if the formula works for n=2 (it does) then by step 2, it works for n=2+1=3

if the formula works for n=3, then by step 2, it works for n=4

and we keep going forever.... because the formula works for n+1, we can prove it works for each number in turn, forever...

Answer 15

Oh, just got it. Had to stare at it for a while. Thanks so much! :DDD God bless you!

Answer 16

Here is more background
http://www.khanacademy.org/math/precalculus/seq_induction/proof_by_induction/v/proof-by-induction

It is a tricky idea.