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'Summing Squares' printed from https://nrich.maths.org/
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:
$$ 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6 $$