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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?
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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.

Vernon, Sep 05 2016

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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!)
Vernon, Sep 05 2016
  

       Ha!
Guessed the author from the title alone.
  

       On the first read through I didn't realise you meant square roots so I kept picturing squares on a game board.   

       ...and just in case you hadn't noticed, you started a sentence with "But plug".
: ]
  

       ^ I wasn't going to mention that...
not_morrison_rm, Sep 05 2016
  

       Somebody had to let him know he'd made an inyouendoh.   
      
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