Prove $\frac{11n^3 + 25n}{6}$ is an integer

Question

Prove by induction that for every integer n, $\frac{11n^3 + 25n}{6}$ is an integer (i cant post the actual given expression since it might cause academic offence). I have only worked with proofs with natural numbers so induction was quite linear. For integers how do i go about showing that it is an integer. I tried cases where n = 1, 2, 3, -1, -2, -3 and it is all true but i dont know how to start it. Any insight would be appreciated. Thank you for your time.

Answer 1

For your particular expression, $$f(n)=\frac{11n^3+25n}{6}.$$

We have the property that $f(-n)=-f(n)$, hence if $f(n)$ is an integer, $f(-n)=-f(n)$ must also be an integer.

Hence, you can focus on proving that for natural number $n$, $f(n)$ is an integer.

For base case, do it for $P(0)$.

Induction step: show that $P(n) \implies P(n+1)$

If you insist: show that $P(n) \implies P(n-1)$ as well. For this particular problem, once you find a relationship between $f(n+1)$ and $f(n)$, both directions should be similar.

I would actually recommend understanding the expression using modulo $6$ arithmetic besides induction.

Answer 2

Rewrite:

$$A=11n^3+25n=12n(n^2)-n(n-1)(n-2)$$

$A_1=(n-1) n (n+1)=6 k$

Because it is the product of three consecutive numbers.So A is always divisible by 6,so $A/6$ is integer. To show that by induction put $n+1 $ you get:

$A_1=n(n+1)(n+2)$

which is again a product of three consecutive numbers and is a multiple of 6, that is $A/6$ is integer for both n and n+1 so it is integer for every n.

Answer 3

Start anywhere.

If $n = 59$ then $\frac {11n^3 +25n}6 = \frac {11*59^3 + 25*59}6 = \frac {2260644}6 =376774$.

Now prove by induction that we can "go up". If $\frac {11n^2 + 25n}6$ is an integer then prove $\frac {11(n+1)^2+ 25*(n+1)}6$ is an integer.

So $\frac {11(n+1)^3 + 25(n+1)}6 = \frac {11(n^3 + 3n^2 + 3n + 1) + 25n + 25}6=$

$\frac {11n^3 + 25n + (33n^2 + 33n + 36)}6= \frac {11n^3+25n}6 + \frac {11n^2 + 11n + 12}2=$

$\frac {11n^3 +15n}6 + 11\frac {n(n+1)}2 + 6$.

Now $\frac {11n^3 + 15n}6$ is an integer. Ad as either $n$ or $n+1$ is even $11\frac {n(n+1)}2$ is an integer and $6$ is an integer. SO the sum is an integer.

So you have proven if $n =59$ or $n$ is an integer that can be reached from $59$ by adding $1$ any number of times then $\frac {11n^3 + 25n}6$ is an integer.

Now prove by induction that we can "do down. If $\frac {11n^2 + 25n}6$ is an integer then prove $\frac {11(n-1)^2+ 25*(n-1)}6$ is an integer.

So $\frac {11(n-1)^3 + 25(n-1)}6 = \frac {11(n^3 - 3n^2 + 3n - 1) + 25n - 25}6=$

$\frac {11n^3 + 25n + (-33n^2 + 33n - 36)}6= \frac {11n^3+25n}6 + \frac {-11n^2 + 11n - 12}2=$

$\frac {11n^3 +15n}6 -11\frac {n(n-1)}2 - 6$.

Now $\frac {11n^3 + 25n}6$ is an integer. And as either $n$ or $n-1$ is even $11\frac {n(n-1)}2$ is an integer and $6$ is an integer. SO the sum is an integer.

Wo we have proven if $n=59$, or if $n$ can be reached from $59$ by repeatedly adding $1$, or if $n$ can be reached from $59$ by repeatedly subtracting $1$, we have proven our result.

And that covers every integer.

....

We could simplify it.

Start with Base case $n=0$, have the induction step go "both ways" and prove for if true for $n$ then is true for $n\pm 1$.

That's do it.

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or note: $\frac {n^3 + 25n}6$ is an integer if and only if $\frac {(-n)^3 + 25(-n)}6 = -\frac {n^3 + 25n}6$ is an integer.

So we don't have to prove for all $n$. Just all natural $n$.