Growing Surprises

Thank you to everybody who submitted solutions for this problem. Here's how some of you thought about the problem:

Tor, from Harbinger School, UK,  and Edi, Joseph, Vladimir and Alastair, from The British International School of Shanghai Puxi, China, had a similar approach to the first part of the problem:

I first found the total number of squares in each pattern, and then computed the differences between them. So, the totals are: 8, 16, 24, 32, ... Therefore, the difference between any 2 neighbours is 8, so the $n^\text{th}$ pattern has $8n$ squares.

Then I can easily work out the number of squares in the $ 20^\text{th}$ pattern (there are $8$ x $20 = 160$ squares) or in the $50^\text{th}$ pattern ($8$ x $50 = 400$ squares).

Abby and Rebecca, from the British International School, Shanghai, Puxi campus, and Rosie Coward, from Wycombe High School, England, worked out the formula for the second pattern:

Taking the next sequence, we realised that these were the same shapes, but filled in, rather than hollow. We counted each number of squares from the first patterns, and worked out the differences between each neighbour. Then we did that again, with the new sequence:

So the second difference is 8. Since to find the multiplier (coefficient) of the squared number, you must divide the second difference by two, we found the first term, $4n^2$. Comparing this new sequence $(4, 16, 36, 64)$ to our original one gives a difference of $4n$. Therefore, the number of squares in the $n^\text{th}$ pattern is $4n^2 + 4n$.

From there we found the next 3 terms (120,168, 224), and then found the 20th term (1680)
and the 50th (10200).

Tim, from Gosforth Academy, UK, used a geometric approach to the problem:

Each shape consists of two $2n-1$ sides (the left and right ones, in green) and two $2n+1$ sides (the top and bottom ones, in pink). The diagram shows the $n=6$ case:

Therefore, the total number of squares in the $n^\text{th}$ pattern is $2(2n+1) + 2(2n-1) = 8n$.

Interestingly, if we fit the pieces into each other, we get the solid shapes shown in the second sequence:

Michael and Lera, from Harbinger Primary, UK, expressed it in a slightly different way. 

In the second sequence, each pattern is obtained from the previous one in the sequence, plus a contour from the first sequence:

Nia from School No. 97, Bucharest, Romania, created the next two patterns in each of the first two sequences, and gave a slightly different gometric interpretation:

The next two patterns (the fifth and sixth) in the first and second sequences are:

 

We divide $4$ by $2$ to get the coefficient of $n^2$: so we have $2n^2$. Then we subtract $2n^2$ from the original sequence for each term and we get $6, 10, 14, 18, ...$. The difference between two consecutive term is $4$, so $n$ has a coefficient of $4$. Therefore, the $n^\text{th}$ pattern of the sequence has $2n^2 + 4n + 2$ squares. 

As we looked at them, we realized the reason why one was linear while the other one was quadratic. The first one was counting squares on the contour, which is simply finding the perimeter. The second one was finding the area of an almost full square, so it should have an $n^2$ term. In other words, the perimeter has only one dimension, while areas have two dimensions.

Nia from School No. 97, Bucharest, Romania, approached this part of the problem geometrically:

By moving the yellow squares from the picture below diagonally up, we can see that the first pattern is obtained from the contour of an $2+n$ by $2n+1$ rectangle. The diagram shows the $n=4$ case.

                              
So the number of squares in the pattern is the same as in the contour of an $n+2$ by $2n+1$ rectangle, and hence, we can count the squares: