Question 848493
Geometric series with a = first term  and r = common ratio and n = # of terms

Finite Geometric series   {{{a*((1-r^n)/(1-r))}}}

Infinite Geometric series  {{{a/(1-r)}}}

{{{S_7}}} n = 7, a = 4, r = 1/3    {{{4*((1-(1/3)^7)/(1-(1/3))) = 4372/729 }}}


{{{S_inf}}} a=4, r=1/3   {{{4/(1-(1/3))}}} =  6

{{{6- (4372/729) =  highlight(2/729)}}}  

Alternative method:

Calculating {{{a/(1-r)  - a*((1-r^n)/(1-r))}}}

{{{a/(1-r)  - (a-ar^n)/(1-r))}}}

{{{ar^n/(1-r)}}}


{{{(4*(1/3)^7)/(1-(1/3))  = highlight(2/729)}}}