The nth even and odd number

Algebraic expressions have parity too, and it need not be fixed. Take 3n + 4. For n = 3 it is 13, odd. For n = 8 it is 28, even. For n = 10 it is 34, even. So this expression is sometimes odd and sometimes even.

Can you come up with an expression that is ALWAYS even? Some examples are 100p — always a multiple of 100, which is even — and 48w − 2. Try to find more, and then try to find expressions that are always odd.

Now a sharper question. The expression 6k + 2 evaluates to 8, 14, 20, … for k = 1, 2, 3, …. Those are all even, but many even numbers are missing. Is there an expression that lists ALL the even numbers? The hint is that every even number has a factor 2.

You already found the nth term of the multiples of 4 to be 4n. So what is the nth term for multiples of 2 — that is, the nth even number? It is 2n.

Now for the odd numbers. What is the 100th odd number? Start with a question you can answer: what is the 100th EVEN number? It is 2 × 100 = 200.

Does that help? Line the two sequences up term by term. Even numbers: 2, 4, 6, 8, 10, 12, … Odd numbers: 1, 3, 5, 7, 9, 11, … At every position, the odd number is exactly one less than the even number. So the 100th odd number is 200 − 1 = 199.

Write the method down as a formula. To find the odd number at a given position: first find the even number at that position, which is 2 times the position number; then subtract 1. In expressions, those two steps are 2n and 2n − 1. So 2n gives the nth even number and 2n − 1 gives the nth odd number.