# Steps to find the general formula for the series

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!

• 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? – user228113 Dec 18 '17 at 23:25
• Yes!, right like that, the thing is to find the formula. – Hamilton Tobon Dec 18 '17 at 23:27

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*}