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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

Age 14 to 16 Challenge Level:

You may wish to make six copies of each square prism below to aid you in visualising the problem that follows.
1 by 1 by 1, 2 by 2 by 1, 3 by 3 by 1 cuboids
If I take six of the red cubes I can make this 3 by 2 by 1 cuboid:
3 by 2 by 1 cuboid
I want to surround this with blue cubes to make the 5 by 3 by 2 cuboid below:

5 by 3 by 2 cuboid
Could I build this cuboid by adding on six blue square prisms like the one above without splitting any of them?
If so, describe how to do it. If not, explain why not, and describe how to arrange the blue square prisms so that as few as possible need to be split.

I want to surround this with pink cubes to make a 7 by 4 by 3 cuboid. (See Hint for a picture of this cuboid.)
Could I build it by adding on six pink square prisms like the one above? Do I need to split any this time?

So far, we've discovered that
$$6 \times 1^2 = 3 \times2 \times1$$
$$6\times(1^2+2^2) = 5\times3\times2$$
$$6\times(1^2+2^2+3^2) = 7\times4\times3$$

Can you describe a method for forming the next cuboid in the sequence by adding on six square prisms? You will only need to split one of your square prisms each time. Does the sequence continue forever?

Work out the dimensions of the $n^{th}$ cuboid.
Can you explain how to make the $(n+1)^{th}$ cuboid from it? What are the dimensions of the $(n+1)^{th}$ cuboid?

How could these ideas be used to calculate the sum $1^2+2^2+3^2+...+n^2$ quickly for any value of $n$?