Magic of numbers on getting squared

 For the difference of consecutive squares, we start with two consecutive numbers \( n \) and \( n+1 \), and the difference of their squares is:

\[

(n+1)^2 - n^2

\]

Expanding \( (n+1)^2 \):

\[

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

\]

So, the difference between the squares of consecutive numbers is:

\[

(n+1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1

\]

### Let's calculate this for \( n = 1 \) through \( n = 10 \):

- For \( n = 1 \):

  \[

  2(1) + 1 = 3

  \]

- For \( n = 2 \):

  \[

  2(2) + 1 = 5

  \]

- For \( n = 3 \):

  \[

  2(3) + 1 = 7

  \]

- For \( n = 4 \):

  \[

  2(4) + 1 = 9

  \]

- For \( n = 5 \):

  \[

  2(5) + 1 = 11

  \]

- For \( n = 6 \):

  \[

  2(6) + 1 = 13

  \]

- For \( n = 7 \):

  \[

  2(7) + 1 = 15

  \]

- For \( n = 8 \):

  \[

  2(8) + 1 = 17

  \]

- For \( n = 9 \):

  \[

  2(9) + 1 = 19

  \]

- For \( n = 10 \):

  \[

  2(10) + 1 = 21

  \]

So, the sequence of differences between consecutive squares is:

\[

3, 5, 7, 9, 11, 13, 15, 17, 19, 21

\]

### Observations:

- The differences between the squares of consecutive numbers follow a constant pattern of \( 2n + 1 \), which means the differences increase by 2 at each step.

- This leads to a simple arithmetic sequence: \( 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 \), where the difference between each pair of consecutive terms is always 2.

In this case, the differences between consecutive squares form a straightforward, linear pattern.