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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?
 (+2) [vote for, against]

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

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".
: ]
 — 2 fries shy of a happy meal, Sep 05 2016

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