Question 948070

In the sequence {{{3}}}, {{{9}}}, {{{27}}}, {{{81}}} ...

first term  is {{{3}}}
second term  is {{{9}}}=>{{{3^2}}}
third term  is {{{27}}}=>{{{3^3}}}
fourth term  is {{{81}}}=>{{{3^4}}}

more terms
{{{3}}}, {{{9}}}, {{{27}}}, {{{81}}}, {{{243}}}, {{{729}}}, {{{2187}}}, {{{6561}}}, {{{19683}}}, {{{59049}}}, {{{177147}}}, {{{531441}}}, ...

so, general formula is:

{{{a[n] = 3^n}}} (for all terms given)

then {{{a[100] = 3^100}}}

{{{a[100] =515377520732011331036461129765621272702107522001}}}-515 quattuordecillion (a cardinal number represented in the U.S. by 1 followed by {{{45}}} zeros)