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Odd number | Even number | |
Odd number | Odd | Even |
Even number | Even | Even |
If we take three consecutive numbers and multiply them together, the resulting number is always divisible by three. This should be obvious since if we have any three consecutive numbers, one of them will be divisible by 3.
Now we are ready to demonstrate that ${n}^{2}+n$ is even for all $n\in {Z}$ . If we factorise this expression we get: $n(n+1)$ . If $n$ is even, than $n+1$ is odd. If $n$ is odd, than $n+1$ is even. Since we know that if we multiply an even number with an odd number or an odd number with an even number, we get an even number, we have demonstrated that ${n}^{2}+n$ is always even. Try this for a few values of $n$ and you should find that this is true.
To demonstrate that ${n}^{3}-n$ is divisible by 6 for all $n\in {Z}$ , we first note that the factors of 6 are 3 and 2. So if we show that ${n}^{3}-n$ is divisible by both 3 and 2, then we have shown that it is also divisible by 6! If we factorise this expression we get: $n(n+1)(n-1)$ . Now we note that we are multiplying three consecutive numbers together (we are taking $n$ and then adding 1 or subtracting 1. This gives us the two numbers on either side of $n$ .) For example, if $n=4$ , then $n+1=5$ and $n-1=3$ . But we know that when we multiply three consecutive numbers together, the resulting number is always divisible by 3. So we have demonstrated that ${n}^{3}-n$ is always divisible by 3. To demonstrate that it is also divisible by 2, we can also show that it is even. We have shown that ${n}^{2}+n$ is always even. So now we recall what we said about multiplying even and odd numbers. Since one number is always even and the other can be either even or odd, the result of multiplying these numbers together is always even. And so we have demonstrated that ${n}^{3}-n$ is divisible by 6 for all $n\in {Z}$ .
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