Question 1070873
{{{81=9^2=(3^2)^2=3^(2*2)=3^4}}}
Mental math says that the common ratio is {{{r=3}}} .
With formulas, id {{{b[1]}}} is the first term,
and {{{r}}} is the common ratio,
{{{b[n]=b[1]*r^(n-1)}}} , so
{{{b[5]=b[1]*r^4}}} is {{{81}}} , the fifth term,
{{{b[2]=b[1]*r}}} is {{{3}}} , the second term,
{{{b[5]/b[2]=b[1]*r^4/(b[1]*r)}}} ,
{{{b[5]/b[2]=r^3}}} , and substituting
{{{81/3=r^3}}} ,
{{{3^4/3=r^3}}} ,
{{{3^3=r^3}}} , and {{{r=3}}} .
The sum of {{{n}}} terms starting from a term {{{b}}}
in a geometric series with common ratio {{{r}}} , is
{{{b(r^n-1)/(r-1)}}} .
We are skipping the first {{{4}}} terms of a geometric series with {{{r=3}}} ,
and adding up the next {{{n=10-4=6}}} terms,
starting from fifth term {{{b=81}}} and adding up to the tenth term,
so the sum is
{{{81(3^6-1)/(3-1)=81(729-1)/2=81*728/2=81*364=highlight(29484)}}}
You may think that I need to find the first term of that series,
then calculate the sum of the first 10 terms,
next calculate the sum of the first 4 terms,
and finally subtract one sum from the other.
That is inefficient.
Who can say that I am not allowed to think of my own series,
which happens to have the same {{{r}}} as the series in the problem,
and the same terms, except for missing the first 4?