Show that a function f(n)= 2n^3+n^2+6n+3 always produces a number that is divisible by an odd number greater than 1, for any natural number, n.

Well, if n is a whole number, then 2n will always be even.

When you add 1 to any even, the result will be odd.
So 2n+1 is always odd
by writing it as
2n^3+n^2+6n+3
= (2n+1)(n^2 + 3)

I have shown that one of the factors, namely (2n+1), will always be an odd number, thus
2n^3+n^2+6n+3
is always divisible by an odd number

let's factor it ...

2n^3+n^2+6n+3
= n^2(2n+1) + 3(2n+1)
= (2n+1)(n^2 + 3)

what kind of a number is 2n+1 ??

To show that a function f(n) always produces a number divisible by an odd number greater than 1, we need to prove that the function is divisible by at least one odd number greater than 1 for any natural number n.

Let's start by analyzing the terms of the function f(n) = 2n^3 + n^2 + 6n + 3:

Term 1: 2n^3
Term 2: n^2
Term 3: 6n
Term 4: 3

Notice that the first three terms, 2n^3, n^2, and 6n, are all divisible by n (since n is a common factor). Therefore, we can factor out an n from the first three terms:

f(n) = n(2n^2 + n + 6) + 3

Now, let's focus on the expression inside the parentheses, 2n^2 + n + 6. In order to prove the divisibility by an odd number greater than 1, we can observe the expression modulo an odd number. Let's consider an odd number k, where k > 1.

Take the expression modulo k:

(2n^2 + n + 6) mod k

We can now simplify the expression by taking the modulo of each term individually:

(2n^2 mod k + n mod k + 6 mod k) mod k

Since k is an odd number greater than 1, the modulo operation with k will result in an odd number. Also, any odd number plus an odd number will always be even, and any even number plus an even number will also be even.

Hence, we have:

(2n^2 mod k + n mod k + 6 mod k) mod k = (even + odd + even) mod k = even mod k

So, the expression (2n^2 + n + 6) mod k always evaluates to an even number when k is an odd number greater than 1.

Now, let's substitute this result back into the original function:

f(n) = n(2n^2 + n + 6) + 3 = (even) + 3

No matter what value of n we choose, (even) + 3 will always be an odd number. Therefore, we can conclude that the function f(n) = 2n^3 + n^2 + 6n + 3 always produces a number that is divisible by an odd number greater than 1 for any natural number n.

I do not know.