h a l f b a k e r yTrying to contain nuts.
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Series of Squares Formula
What's the square that is reached, in a series that starts at a specified Origin-number and skips a particular Interval, N times? | |
S = (I*N)^2 + 2*I*N*O + O^2
where O is the Origin, I is the Interval, and N is a Number of
Increments in the series, and S is the Square. For example, if
the Origin is 7, and the Interval is 6 at a time, and you want N
to be the number at the 8th increment of the series...
7^2=49 --the
Origin is always the Zeroth member of the series!
(Think: "Zero Increments".)
13^2=169 --first member after the Origin, Incrementing by
six
19^2=361, and you have six more Increments to go to reach the
number at the eighth one.
But plug the numbers appropriately into the formula, and:
S=(6*8)^2 + 2*6*8*7 + 7^2 =
48^2 + 672 + 49 =
2304 + 721 =
3025 (the number at the 8th Increment of the above series is
55, and 3025=55^2)
This is HalfBaked because while it certainly works perfectly,
I'm
not sure what anyone would actually use it for.
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I suppose some wag might say:
S = [(I*N) + O]^2
and be correct; my formula is basically what you get if you
algebraically do the squaring of [(I*N) + O]. But that's not
how I originally derived it; I derived it from studying a
bunch of sequences of numbers and their squares! (On the
other hand, the simplification proves the other formula is
valid!) |
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Ha!
Guessed the author from the title alone. |
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On the first read through I didn't realise you meant square roots so I kept picturing squares on a game board. |
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...and just in case you hadn't noticed, you started a sentence with "But plug".
: ] |
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^ I wasn't going to mention that... |
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Somebody had to let him know he'd made an inyouendoh. |
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