Question 1186917


We know the first term is {{{4}}}, so 
{{{a[1]=4}}}

Each term is {{{3}}} times bigger than the last, meaning we have  {{{a[1]r^(n-1)}}}, with {{{r=3}}} 

So, we know the series follows {{{4(3)^(n-1)}}}.

Foe a geometric series:

{{{S[n]=a[1]((1-r^n)/(1-r))}}}

we have: {{{a[1]=4}}} and {{{r=3}}} 

We need {{{n}}} for the last term:

{{{4(3)^(n-1)=8748}}}

{{{3^(n-1)=2187}}}.......since {{{2187=3^7}}}

{{{3^(n-1)=3^7}}}........since bases same, equal the exponents

{{{n-1=7}}}

{{{n=8}}}

{{{S[8]=4((1-3^8)/(1-3))}}}
{{{S[8]=4((-6560)/(-2))}}}
{{{S[8]=4(3280)}}}
{{{S[8]=13120}}}