Question

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 1

1. prove it is true for n=1
2. assume n=k
3. prove that n=k+1 is true as well


so

1.
$\frac{n^3+2n}{3}$=
$\frac{1^3+2(1)}{3}$=
$\frac{1+2}{3}$=1
we got a whole number, true


2.
$\frac{k^3+2k}{3}$
if everything clears, then it is divisble


3.
$\frac{(k+1)^3+2(k+1)}{3}$ =
$\frac{(k+1)^3+2(k+1)}{3}$ =
$\frac{k^3+3k^2+3k+1+2k+2)}{3}$=
$\frac{k^3+3k^2+5k+3)}{3}$
we know that if z is divisble by 3, then z+3 is divisble b 3
also, 3k/3=a whole number when k= a whole number

$\frac{k^3+2k}{3} + \frac{3k^2+3k+3}{3}$=
$\frac{k^3+2k}{3} + k^2+k+1$=
since the k²+k+1 part cleared, it is divisble by 3

we found that it simplified back to $\frac{k^3+2k}{3}$

done

Answer 2

Answer:

We have to use the 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.

• for n=1

$n^3+2n=1+2=3$ is divisible by 3.

Hence, the statement holds true for n=1.

• Let us assume that the statement holds true for n=k.

i.e. $k^3+2k$ is divisible by 3.---------(2)

• Now we will prove that the statement is true for n=k+1.

i.e. $(k+1)^3+2(k+1)$ is divisible by 3.

We know that:

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

and $2(k+1)=2k+2$

Hence,

$(k+1)^3+2(k+1)=k^3+1+3k^2+3k+2k+2\\\\(k+1)^3+2(k+1)=(k^3+2k)+3k^2+3k+3=(k^3+2k)+3(k^2+k+1)$

As we know that:

$(k^3+2k)$ was divisible as by using the second statement.

Also:

$3(k^2+k+1)$ is divisible by 3.

Hence, the addition:

$(k^3+2k)+3(k^2+k+1)$ is divisible by 3.

Hence, the statement holds true for n=k+1.

Hence by the mathematical induction it is proved that:

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