Question 969177
<pre>
That's a geometric sequence.

{{{a[n] = a[1]r^(n-1)}}} where {{{a[1]=4}}} and {{{r=-3}}}

So the nth term is

{{{a[n] = 4*(-3)^(n-1)}}}

{{{sum((4*(-3)^(n-1)),n=4,15)}}}

Edwin</pre>