how do i prove by mathematical induction?

Question

n^3 + 2n is divisible by 3 for all n = positive integers

Answer 1

We need to prove 2 things in mathemetcal induction.
it is true for the lowest one n = positive so n = 1

n^3+2n= 1+2 = 3

which is true.

Now we need to prove that if it is true for n then for n+1
let f(n) = n^3+2n
f(n+1) = (n+1)^3 + 2(n+1)
f(n+1) - f(n) = (n+1)^3+2(n+1) -(n^3+2n)
= n^3 + 3n^2 + 3n + 1 + 2n + 2 - n^3 - 2n
= 3n^2 + 3n + 3 or 3(n^2+n+1)
which is dvivisible by 3
so f(n+1) is divisible by 3 if f(n) is which is true

we have proved both parts
QED

Answer 2

n^3+2n

if n is an even integer, than n=2 is also an even integer
(n+2)^3+2(n+2)
n^3+6n^2+12n+12+2n+4
subtract n^3+2n leaves
6n^2+12n+12=6(n^2+2n+2). If n is an integer, this expression must be divisible by 3. therefore if this is valid for 1 even integer, it is will be valid for all since we can just keep adding 2 to to an even integer to get all other even integers.

let n=2
2^3+2*2=8+4=12 which is divisible by 3.
Therefore n^3+2n is divisible by 3 for n being any even integer.

since for n=1, 1^3+2*1=3, the premise is true for n= any integer, odd or even.

Answer 3

To use induction, math_kp's solution is good. But this problem does not require induction.

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

Hence, n^3+2n is divisible by 3 if and only if (n-1)(n)(n+1) is divisible by 3. (Note that 0 *is* divisible by 3, since 0=3*0.) But n-1, n, and n+1 are three consecutive integers, so one of them *must* be divisible by 3. Hence, n^3+2n is always divisible by 3.

My point is, just because the statement says "for all positive integers", induction is not always necessary and may not even be the easiest method of proof.

Answer 4

the case n=1 is true right?
Now keep that aside and state that the statement is true for n=p for some p,
then p^3 +2p is divisible by 3
now replace p by p+1, then
(p+1)^3 + 2(p+1) arises, now show this interms of the expression for p and a number or term evidently divisible by 3.
Then for n=p+1 the result holds.Since we already have a truth in the onset of the set of positive integers, the result is true for all positive integers.
QED

Answer 5

First prove that this is true for n = 1. Next, assume it's true for an arbitrary n. Finally, use these two (the proof when n = 1, along with your assumption) to prove that it's true for n+1.