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The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.

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Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?

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Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?

Summing Squares

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

We received two very detailed solutions for this problem, one by Rajeev from Fair Field Junior School and one by Furkan from Amasya Science and Arts Centre.

Both correctly concluded that in order to build the 5 by 3 by 2 cuboid using the blue square prisms they would have to split one of the square prisms. Here is Furkan's pictorial representation of how to build the cuboid.

Furkan continues:

We can also surround the above cuboid by six pink prisms. Again we have to split one prism as seen in the diagram below.

Rajeev noticed this pattern continued for the next 4 cuboids in the sequence.

We discovered that:

$$ 6 \times 1^2 = 3 \times 2 \times 1$$
$$ 6 \times (1^2 + 2^2) = 5 \times 3 \times 2 $$
$$ 6 \times (1^2 + 2^2 + 3^2) = 7 \times 4 \times 3 $$

Rajeev then reasoned:

The method for forming the next cuboid is adding 6 square prisms and the dimensions of the prisms are the shortest dimension of the new cuboid. For example in the $7 \times 4\times 3$ cuboid the shortest dimension is 3 so the prism is 3 by 3 by 1. To form the next cuboid surrounded say by 6 yellow square prisms the dimensions of the prisms would be 4 by 4 by 1 and the dimension of the cuboid would be 9 by 5 by 4. The formula for the number of cubes to add on is $6 \times (n+1)^2$ i.e. 6 square prisms of side n+1.

The method for finding out the number of cubes used is $(2n+1) \times (n+1) \times n$ where n is the shortest side.

I also noticed that in each cuboid the width plus the height of the cuboid equals the length. So for example in the cuboid where 12 is the shortest side the cuboid has dimensions $ (2n+1) \times (n+1) \times n = 25 \times 13 \times 12 $.

The dimension for the $n^{th}$ cuboid is n, n+1 and 2n+1. And for the $(n+1)^{th}$ cuboid it is n+1, n+2 and 2n+3.

So we can write the $n^{th}$ cuboid as
$$6 \times (1^2 + 2^2 + ... + n^2) = (2n+1) \times (n+1) \times n$$

Thes ideas can be used to sum the series $ 1^2 + 2^2 + 3^2 + ... + n^2 $ quickly for any value of n using the formula:

$$ 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6 $$