Question 1197507
For a geometric series, the n-th partial sum is:
{{{ S[n] = a[1]*((1-r^n)/(1-r)) }}}


This problem presents a geometric sequence with {{{a[1] = 16}}} and {{{r=3/4}}}.
The infinite sum exists because {{{abs(r)<1}}}, and is the limit of {{{S[n]}}} as n goes to infinity:

Using "Lim" to denote "Limit as n goes to infinity":

 Lim ({{{ S[n] }}}) =  Lim ({{{ 16*(1-(3/4)^n)/(1-(3/4)) }}} )
    
= Lim  ( {{{ 16 * 4 * (1-(3/4)^n) }}} )

= ( 16 * 4 * 1 )

= 64