Don't fall in

By separating the sinkhole into a pattern of alternating colours, we can see each pattern is made of squares in a square formation, with the number of red squares equal to the year squared and the number of blue squares the year minus 1 squared.

Since the sinkhole starts with $1$ unit and its width grows by $2$ units every year, the sinkhole is $2n - 1$ units across for year $n$ .

When the year $n$ is odd, the sinkhole can be partitioned by its center $n$ x $n$ square, leaving $4$ staircase shapes left that have a base of $n - 2$ and a height of $\frac{1}{2}(n - 1)$ that can be rearranged into a second $(n - 1)$ x $(n - 1)$ square, as shown below.

When the year $n$ is even, the sinkhole can be partitioned by its center $(n - 1)$ x $(n - 1)$ square, leaving $4$ staircase shapes left that have a base of $n - 1$ and a height of $\frac{1}{2}n$ that can be rearranged into a second $n$ x $n$ square, as shown below.

Either way, the sinkhole area for year n is always $(n - 1)^2 + n^2$ , the sum of the squares of two consecutive integers.

Let $A_n$ be the area of the sinkhole in year $n,$ where $n \geq 1.$ We will prove by induction that $A_n = (n - 1)^2 + n^2.$ The base case, $n = 1,$ is $A_1 = 1 = 0^2 + 1^2.$ Now, assume that the formula holds for $n = k.$ Note that after year $k,$ the sinkhole grows by an additional $4k$ square meters, since each outer edge (northeast, northwest, southeast, and southwest) has $k$ unit squares. Thus, we have

$\begin{aligned} A_{k + 1} &= A_k + 4k \\ &= (k - 1)^2 + k^2 + 4k \\ &= k^2 - 2k + 1 + k^2 + 4k \\ &= k^2 + 2k + 1 + k^2 \\ &= k^2 + (k + 1)^2. \end{aligned}$

This completes our inductive step, so we conclude that yes , the area of the sinkhole is always the sum of two consecutive perfect squares.

We will prove this statement true through induction. Let the area of the sinkhole on year $y$ be $S_y$ . We have a few base cases given, so we must prove that the area will be a sum of two consecutive squares for all years greater than $4$ .

Each year, the sinkhole retains its area from the previous year (in black) but also grows additional squares. The sinkhole always grows $4$ blue squares each year, but the number of red squares it grows increases by $4$ each year. Because of this, we have the following recursive relationship:

$S_1 = 1$ $S_y = S_{y-1}+ \color{#3D99F6}4 \color{#333333}+ \color{#D61F06}4(y-2) \color{#333333} = S_{y-1}+4y-4, \text{ }y>1$

Now assume that on some year $k$ , $S_k = (k-1)^2+k^2$ .

$S_{k+1} = S_k+4(k+1)-4 = (k-1)^2+k^2+4k = 2k^2+2k+1 = k^2+(k^2+2k+1) = k^2+(k+1)^2$

Therefore, the area of the sinkhole on any given year is the sum of two consecutive perfect squares.

The area of the sinkhole for year $n$ may be expressed as $\begin{aligned} A_n &= 1 + \sum_{k = 1}^{n}4(k-1) \\ &= 1 + 4\sum_{k = 1}^{n}(k-1) \\ &= 1 + 4 \frac{n(n-1)}{2} \\ & = 1 + 2n(n-1) \\ & = 1 + 2n^2 -2n \\ & = \underbrace{n^2 - 2n + 1}_{(n-1)^2} + n^2 \\ & = (n-1)^2 + n^2 \quad \square \end{aligned}$

If we consider the number of squares in each "column" of the sinkhole we see that they will always be odd numbers, as they begin with 1 and increase by a multiple of 2 each time: 1, 3, 5, 7 etc.

Consequently, reading across the sinkhole and adding up the squares in each column, we find they are composed of sequences of odd numbers, e.g. Year 3 is 1 3 5 3 1. They are also symmetrical, with the numbers on either side of the middle column being the same. Consequently we can consider them to be made of two sequences with n and n+1 odd numbers, e.g. 1 3 5 and 1 3 in the case of Year 3.

As the first n odd numbers sums to n^2, this means that the sinkholes will always be composed of n^2 + (n+1)^2 squares

The annual increase increases by 4 units. Therefore the area is a quadratic function of time, with leading coefficient $\tfrac12\cdot 4$ : $A_n = 2 n^2 + b n + c.$ WIth $A_1 = 1$ and $A_2 = 5$ , it is easy to see that $b = -2$ and $c = 1$ . Now $A_n = 2 n^2 - 2n + 1 = (n^2 - 2n + 1) + n^2 = (n-1)^2 + n^2.$ $\boxed{\text{Yes}}$ , the area is always the sum of the squares of two consecutive integers.

By considering the first example you can notice the following pattern for f(n) modelling its area:

$f(1) = 1 + \textbf{0} \cdot 4$ , $f(2) = 1 + \textbf{1} \cdot 4$ , $f(3) = 1 + \textbf{3} \cdot 4$ , $f(4) = 1 + \textbf{6} \cdot 4$ , $f(5) = 1 + \textbf{10} \cdot 4$ .

So we can deduce, that $f(n) = 1 + \binom{n}{2} \cdot 4 = 1 + \frac{n!}{2! (n - 2)!} \cdot 4 = 1 + 2n (n - 1) = 2n^2 - 2n + 1$ .

The pattern which is suggested in the question is: $f(n) = (n-1)^2 + n^2$ . By multiplying that out, we see, that they are indeed equivalent.

Note that the sum of the first $n$ odd integers is always $n^2$ . The top half with the middle has an area equal to the sum of the first $n$ odd integers if you add up all the row lengths, where $n$ is the year number. So, that area is $n^2$ . Similarly, the bottom half has an area equal to the sum of the first $n-1$ odd integers, which is equal to $(n-1)^2$ . So, the sinkhole area is always the sum of the square of two consecutive integers.

In the initial case when area =1then area=0^2+1^2. Let the induction hold good for n cases.Now,for n+1 Area of n+1=(n-1)^2+n^2+4n =n^2+1^2-2n+n^2+4n =n^2+1^2+2n+n^2 =n^2+(n+1)^2. Therefore it hold good for n+1 case also Hence, the answer is yes.....

Always 4 squares more to add, not including the first one. So add 4 squares to first one and then 8 squares to the second one and then 12 square to the third one and then 16 to the fourth one and 20 to the fifth one and 24 to the sixth one and till infinity. x= y+4 where "y" is a sum of squares added to each edge.

0^2 + 1^2 = 1

1^2 + 2^2 = 5

2^2 + 3^2 = 13

3^2 + 4^2 = 25

4^4 + 5^2 = 41

---

1 + 0*4 + 4 = 5

5 + 1*4 + 4 = 13

13 + 2*4 + 4 = 25

25 + 3*4 + 4 = 41

---

1 + 0(4) = 1

1 + 0(4) + 1(4) = 5

1 + 0(4) + 1(4) + 2(4) = 13

1 + 0(4) + 1(4) + 2(4) + 3(4) = 25

1 + 0(4) + 1(4) + 2(4) + 3(4) + 4(4) = 41

---

x^2 + (x+1)^2 = 4(sum(0..x)) + 1

x^2 + (x+1)^2 = 4(x*(x+1)/2) + 1

x^2 + (x+1)^2 = 2x*(x+1) + 1

x^2 + (x+1)^2 = 2x^2 + 2x + 1

---

x^2 + x^2 + 2x + 1 = 2x^2 + 2x + 1

2x^2 + 2x + 1 = 2x^2 + 2x + 1