Question 1064681
<pre><b>
First we get the general term for 2, 5, 8, 11, ... 
which is an arithmetic sequence and has general term 

{{{a[n]=a[1]+(n-1)d}}}

{{{a[n]=2 + (n-1)(3)}}}

{{{a[n]=2+3n-3}}}

{{{a[n]=3n-1}}}

Now we multiply that general term by the sign-alternating 
factor (-1)<sup>n+1</sup>.  We'll call the general term t<sup>n</sup>.

{{{t[n]=(-1)^(n+1)*(3n-1)}}} 

Edwin</pre></b>