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'Square Holed Pattern' printed from https://nrich.maths.org/
One way to proceed is to regard the pattern as four arms, each two
squares wide, with four corner pieces of three squares each. So for
the
nth pattern, we have $4\times
2\times n + 4\times 3 = 8n+12$. For $n = 10$, we need $ 8
\times 10 +12$ i.e. $92$ squares.
Alternatively, it is possible to see the patter as a complete
square with corners and a central square removed. So for the
nth pattern, we have a complete
$(n+4)(n+4)$ square with the four corners and a central $n\times n$
square removed. Hence the number of squares is $(n+4)^2 - n^2 -4 =
8n +12$.