Question 1047528
It's quite obvious from the fact {{{S[30]= 18*((1-(2/3)^30)/(1/3))}}} that

{{{a[1] = 18]}}} and r = 2/3

a.  {{{a[1] = 18]}}}, {{{a[2] = 12]}}}, {{{a[3] = 8]}}}, {{{a[4] = 16/3]}}}, {{{a[5] = 32/9]}}}.


b.  {{{S[1000000] = a[1]/(1-r) - r^1000000/(1-r) =  18/(1-2/3) - (2/3)^1000000/(1-2/3)}}}.

Since 0 < 2/3 < 1, {{{(2/3)^1000000/(1-2/3) = 3*(2/3)^1000000}}} is practically 0.

Thus an estimate of {{{S[1000000]}}} is {{{ 18/(1-2/3) = 54}}}.