Sunday Afternoon Maths LVIII
Factorial pattern
$$1\times1!=2!-1$$
$$1\times1!+2\times2!=3!-1$$
$$1\times1!+2\times2!+3\times3!=4!-1$$
Does this pattern continue?
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Yes. It can be shown by induction:
First it's easy to check that \(1\times1!=2!-1\). Next assume that \(1\times1!+2\times2!+...+k\times k!=(k+1)!-1\). Now we try to show the pattern holds for \(k+1\).
$$1\times1!+2\times2!+...+k\times k!+(k+1)\times(k+1)!$$$$=(k+1)!-1+(k+1)\times(k+1)!$$
$$=(1+k+1)(k+1)!-1$$
$$=(k+2)!-1$$
Hence, by induction, the pattern holds for all \(k\geq1\).
Placing plates
Two players take turns placing identical plates on a square table. The player who is first to be unable to place a plate loses. Which player wins?
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The first player can always win by first placing a plate on the exact centre of the table. Then the first player can copy what the second player does, but rotated 180°, and hence can always place a plate if the second player could.
Extension
What if the two players play on a regular hexagonal table? Or a regular octagonal table? Or a regular pentagonal table? Or a regular \(n\)-gonal table?
