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Look at this series: 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, ... I've spent two days trying to find the formula for the nth term, but it is too difficult to find a way. Can you help me with the steps to get the general formula for the nth term?. If you can express it as a summation, it would be great. Thanks in advance, God bless you!

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  • $\begingroup$ I presume it's something like you start from zero, then add $1$ twice, then add $2$ twice, then add $3$ twice, is it not? $\endgroup$ – user228113 Dec 18 '17 at 23:25
  • $\begingroup$ Yes!, right like that, the thing is to find the formula. $\endgroup$ – Hamilton Tobon Dec 18 '17 at 23:27
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See https://oeis.org/A002620 "Quarter-squares: floor(n/2)*ceiling(n/2). Equivalently, floor(n^2/4)."

Besides the title, there are dozens of formulas, many of which are summations. Feel free to list your favorite formula here.

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  • $\begingroup$ Very, very useful. Thanks so much! $\endgroup$ – Hamilton Tobon Dec 19 '17 at 0:03
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When you find a new sequence of numbers it is always worth searching it on OEIS (as Chris Cultler has done for you in his answer.)

Note that the odd terms are the square numbers & the even terms are the oblong numbers ($2$ times the triangle numbers.) Their generating functions are \begin{eqnarray*} \sum_{i=1}^{\infty} i^2 x^i =\frac{x(1+x)}{(1-x)^3} \\ \sum_{i=1}^{\infty} i(i+1) x^i=\frac{2x}{(1-x)^3} \end{eqnarray*} When you inter-splice the sequences together you get \begin{eqnarray*} \sum_{i=1}^{\infty} a_i x^i =\frac{x(1+x^2)}{(1-x^2)^3} +\frac{2x^2}{(1-x^2)^3} = \frac{x}{(1-x)^2(1-x^2)}. \end{eqnarray*}

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    $\begingroup$ You are the best bro!. Thanks so much. $\endgroup$ – Hamilton Tobon Dec 19 '17 at 0:23

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