Question 850063
This is a quadratic arithmetic series (sometimes known as second difference)

We notice that the second difference is 1.  (that is the difference is going up by 1).

In general if we have a second difference a, then the n^2 term is a/2.

We also know that the first term is 23, so find a_0 which is 21  (23-2).

so in the form an^2 + bn + c we have

{{{(1/2)n^2 + bn + 22}}}

We set this equal to one of our rows. We could use 1, but I will use a different one just to prove a point. We know that a_2 = 26. So plug in 2 and see if we get 26.

{{{ (1/2) * 2^2 + 2b + 21 = 26}}}
{{{ 2 + 2b + 21 = 26}}}
{{{ 2b = 3}}}
{{{ b = 3/2 }}}

So our form is  {{{(1/2)n^2 + (3/2)n +21}}}