Question 1127336
<pre>
     Term         Diff       
{{{ a[1] = -3 }}}
{{{ a[2] = 0 }}}              3
{{{ a[3] = 4 }}}              4    
{{{ a[4] = 9 }}}              5         
{{{ a[5] = 15 }}}             6  <br>

{{{ a[2] = a[1] + 3 }}}
{{{ a[3] = a[2] + 4 = (a[1]+3) + 4 }}}
{{{ a[4] = a[3] + 5 = (a[2] + 4) + 5 = (a[1]+3)+4+5 }}} <br>

The pattern is:
{{{  a[n] = a[1] + 3+4+5 }}} + . . .  + {{{ (n+1) }}} <br>

If we write it a little differently:
{{{  a[n] = a[1] }}} +  (1+2+3 + . . . +  (n+1)) - 3 <br>

[ Then noting the sum of the first m integers: (1+2+ . . . + m )  =   (m)(m+1)/2  ] <br>

we can simplify:
{{{  a[n] = a[1] }}} +  {{{ ((n+1)(n+2))/2 }}} - 3 <br>

Combining {{{ a[1] = -3 }}} with the -3 we introduced :
{{{ blue (  a[n] = -6 + ((n+1)(n+2))/2 ) }}}    n=1,2,3, . . . <br>


Check with given values:
a[1] = -6 + (2)(3)/2 = -6 + 3 = -3
a[2] = -6 + (3)(4)/2 = -6 + 6 = 0
a[3] = -6 + (4)(5)/2 = -6 + 10 = 4<br>
  
You can plug in n=15 to find a[15] = 130.