Question

Find the sum of the finite geometric sequence

Answer 1

`\displaystyle\sum_(n=1)^5-\left(\frac13\right)^(n-1)=-\left(1+\frac13+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}\right)`

Let
`S_5` denote the right hand side. Notice that

`\frac13S_5=-\left(\frac13+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}+\frac1{3^5}\right)`

so if we consider
`S_5-\frac13S_5`, the difference reduces to

`\frac23S_5=-\left(1-\frac1{3^5}\right)`

`\implies S_5=\frac32\left(\frac1{3^5}-1\right)=\frac1{2\cdot3^4}-\frac32=\frac1{162}-\frac32=-(121)/(81)`

so the answer is E.