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.
2n^3+n^2+6n+3
= n^2(2n+1) + 3(2n+1)
= (2n+1)(n^2 + 3)
what's next?
2n is even, so 2n+1 is odd
since n is a natural number, 2n+1 > 1
so, the value is a multiple of an odd number. You're done.
What's next? Well, it seems like you've made a great observation by factoring out a common factor of (2n + 1) from the expression! Now, let's tackle the two factors individually.
First, we have (2n + 1). Notice that (2n + 1) is an odd number because it has a remainder of 1 when divided by 2. So, we're already halfway there! We just need to show that the other factor, (n^2 + 3), is always divisible by an odd number greater than 1.
Now, let's consider the factor (n^2 + 3). To prove that it is always divisible by an odd number greater than 1, let's take a closer look.
When n is even, we can represent it as n = 2k, where k is a positive integer. Substituting this into our expression, we get:
(n^2 + 3) = (4k^2 + 3)
Since 4k^2 is always even, (4k^2 + 3) will always be odd. Therefore, for even values of n, (n^2 + 3) is divisible by an odd number greater than 1.
When n is odd, we can represent it as n = 2k + 1, where k is a positive integer. Substituting this into our expression, we get:
(n^2 + 3) = (4k^2 + 4k + 1 + 3) = (4k^2 + 4k + 4) = 4(k^2 + k + 1)
In this case, (n^2 + 3) is divisible by 4, which is an even number greater than 1. However, since we are specifically looking for odd divisors greater than 1, we can factor out a 2 from this expression:
(n^2 + 3) = 4(k^2 + k + 1) = 2(2k^2 + 2k + 2)
Now, we have shown that (n^2 + 3) is divisible by an even number, which means it is also divisible by an odd number greater than 1.
Combining both factors, we have:
(2n + 1)(n^2 + 3)
Since (2n + 1)(n^2 + 3) is the product of two factors, one of which is always odd, we can conclude that this expression will always produce a number that is divisible by an odd number greater than 1, for any natural number n.
Hope that explanation made you laugh a little while proving the result! If you have any more questions or need further assistance, feel free to ask.
To show 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, we can examine the expression (2n+1)(n^2 + 3) derived from the original function.
First, let's consider the term (2n+1). We know that for any integer n, the expression (2n+1) is always an odd number.
Next, let's look at the term (n^2 + 3). For any natural number n, the expression (n^2 + 3) will always result in an odd number. This is because for any value of n, the square of n (n^2) will be an even number (since every natural number multiplied by itself is even), and adding 3 to an even number will always result in an odd number.
Now, if we multiply an odd number (2n+1) by an odd number (n^2+3), the product will always be an odd number. This is a result of the product of two odd numbers always being odd.
Therefore, we can conclude that for any natural number n, the function f(n) = 2n^3 + n^2 + 6n + 3 always produces a number that is divisible by an odd number greater than 1.
To show that the expression (2n+1)(n^2 + 3) always produces a number that is divisible by an odd number greater than 1, we need to demonstrate that (2n+1) is always odd and (n^2 + 3) is divisible by an odd number greater than 1 for any natural number, n.
Let's start by proving that (2n+1) is always odd.
An odd number is defined as a number that is not divisible evenly by 2. This means that for any value of n, 2n+1 will always be odd, as adding 1 to an even number (2n) will always yield an odd number.
Next, let's examine (n^2 + 3) and show that it is divisible by an odd number greater than 1 for any natural number, n.
For any value of n, n^2 is always a perfect square. The sum of a perfect square and any nonzero constant is also a perfect square. Therefore, (n^2 + 3) is always a perfect square plus 3.
The sum of a perfect square and 3 will always produce an odd number for any natural number, n. To prove this, consider two cases:
Case 1: n is even.
If n is even, then (n^2 + 3) will be the sum of two even numbers (n^2, 3), which is always even. In this case, n^2 + 3 is divisible by the odd number 1.
Case 2: n is odd.
If n is odd, then (n^2 + 3) will be the sum of an odd number (n^2) and an even number (3), which is always odd. In this case, n^2 + 3 is divisible by the odd number 2.
Since we have shown that (2n+1) is always odd and (n^2 + 3) is always divisible by an odd number greater than 1, we can conclude that the function f(n) = 2n^3+n^2+6n+3 will always produce a number that is divisible by an odd number greater than 1, for any natural number, n.