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What is the best mathematical expression for this?

For any pair of consecutive perfect squares, the quantity of integers in the interval between the two squares is equal to twice the square root of the first square.

The difference between the two squares is equal to one more than the gap.

Now, for a sequence of three consecutive perfect squares, the number of integers in the interval between the first and third squares is equal to (twice the square root of the first square) plus one plus (twice the square root of the second square).

The size of the gap between the first and third perfect squares grows as we go from one sequence of three consecutive perfect squares to the next.

How do I set up a mathematical expression which expresses this growth rate?

For example, between $1$ and $9$, there are $7$ integers. Next, between $4$ and $16$, there are $11$ integers.

I understand that the number of integers in each of these intervals is one less than the difference between the first and third squares.

In the first interval, there are $7$ integers, and in the second, $11$. So $7$ and $11$ are the first two terms in a progression.

Is this progression arithmetic? Is it $7,11,15,19 \ldots$ with $7$ being the first term and the common difference being $4$ as my calculations indicate?

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The $n^2 - (n-1)^2 = (2n-1)$. $(n+1)^2$ and $n^2$ is $(n+1)^2-n^2=2n+1$. Thus the gap between $(n+1)^2$ and $(n-1)^2$ is $(2n-1)+(2n+1) - 1 = 4n-1$

So the progression is as you describe.

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It's arithmetic, as you said.

A common formula for the square of natural number $n$ is the sum of the first $n$ odd numbers:

$$n^2 = \sum_1^n (2n-1).$$

So, the difference between $n^2$ and $(n+2)^2$ is $4(n+1)$, which is arithmetic.

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