# Thread: Prove by induction that: n(n^2 + 3n+2) is divisible by 6

1. ## Prove by induction that: n(n^2 + 3n+2) is divisible by 6

Prove by induction that: $\displaystyle n(n^2 + 3n+2)$ is divisible by 6, for n>0

For n =1,
1(1^2 + 3x1 + 2) = 6
true for n =1

assume true for n = k
$\displaystyle k (k^2 + 3k+2) = 6M$

For n = k + 1,
$\displaystyle k [(k+1)^2 + 3(k+1) + 2]$should be divisible by 6

LHS = $\displaystyle k [(k+1)^2 + 3(k+1) + 2]$
= $\displaystyle (k+1) (k^2+5k+6)$
= $\displaystyle k(k^2+5k+6) + k^2 + 5k+6$
= $\displaystyle k(k^2 + 3k +2) + 2k^2 + 4k + k^2 + 5k + 6$
= $\displaystyle 6M + 3k^2 + 9k + 6$
= $\displaystyle 6M + 3(k^2 + 3k +2)$
= $\displaystyle 6M + 3(6M/k)$
= $\displaystyle 6(M+3/k)$

now I'm stuck. That is obviously not divisible by 6 because M + 3/k is not an integer

HELP

2. Originally Posted by differentiate
Prove by induction that: $\displaystyle n(n^2 + 3n+2)$ is divisible by 6, for n>0

For n =1,
1(1^2 + 3x1 + 2) = 6
true for n =1

assume true for n = k
$\displaystyle k (k^2 + 3k+2) = 6M$

For n = k + 1,
$\displaystyle k [(k+1)^2 + 3(k+1) + 2]$should be divisible by 6

Here it must be $\displaystyle (k+1)\left[(k+1)^2+3(k+1)+2\right]$

Tonio

LHS = $\displaystyle k [(k+1)^2 + 3(k+1) + 2]$
= $\displaystyle (k+1) (k^2+5k+6)$
= $\displaystyle k(k^2+5k+6) + k^2 + 5k+6$
= $\displaystyle k(k^2 + 3k +2) + 2k^2 + 4k + k^2 + 5k + 6$
= $\displaystyle 6M + 3k^2 + 9k + 6$
= $\displaystyle 6M + 3(k^2 + 3k +2)$
= $\displaystyle 6M + 3(6M/k)$
= $\displaystyle 6(M+3/k)$

now I'm stuck. That is obviously not divisible by 6 because M + 3/k is not an integer

HELP

.

3. Thank u for the correction, but I still couldn't get it

4. Originally Posted by differentiate
Prove by induction that: $\displaystyle n(n^2 + 3n+2)$ is divisible by 6, for n>0

For n =1,
1(1^2 + 3x1 + 2) = 6
true for n =1

assume true for n = k
$\displaystyle k (k^2 + 3k+2) = 6M$

For n = k + 1,
$\displaystyle k [(k+1)^2 + 3(k+1) + 2]$should be divisible by 6

LHS = $\displaystyle k [(k+1)^2 + 3(k+1) + 2]$
= $\displaystyle (k+1) (k^2+5k+6)$
= $\displaystyle k(k^2+5k+6) + k^2 + 5k+6$
= $\displaystyle k(k^2 + 3k +2) + 2k^2 + 4k + k^2 + 5k + 6$
= $\displaystyle 6M + 3k^2 + 9k + 6$
= $\displaystyle 6M + 3(k^2 + 3k +2)$
= $\displaystyle 6M + 3(6M/k)$
= $\displaystyle 6(M+3/k)$

now I'm stuck. That is obviously not divisible by 6 because M + 3/k is not an integer

HELP

For $\displaystyle n = k$ you have

$\displaystyle k (k^2 + 3k+2) = 6M$

$\displaystyle k(k + 1)(k + 2) = 6M$.

Also, for $\displaystyle n = k + 1$, you actually have

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

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

$\displaystyle = (k + 1)(k^2 + 5k + 6)$

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

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

$\displaystyle = 6M + 3(k + 1)(k + 2)$

Now, also notice that $\displaystyle (k + 1)(k + 2)$ is always even.

So we could write $\displaystyle (k + 1)(k + 2) = 2N$.

So you now have

$\displaystyle 6M + 3(k + 1)(k + 2) = 6M + 3(2N)$

$\displaystyle = 6M + 6N$

$\displaystyle = 6(M + N)$

which is divisible by 6.

5. Thanks a bunch Prove It. You're a genius

6. Just a note - you may prove this in another, easier way, by noting that $\displaystyle n(n^2+3n+2)=n(n+1)(n+2)$ and by using the fact that out of any three consecutive integers, at least one must be divisible by two and another by three, therefore their product has 2*3=6 as a factor and you're done.