[tex]\displaystyle\sum_{n=1}^5-\left(\frac13\right)^{n-1}=-\left(1+\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}\right)[/tex]
Let [tex]S_5[/tex] denote the right hand side. Notice that
[tex]\dfrac13S_5=-\left(\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}+\dfrac1{3^5}\right)[/tex]
so if we consider [tex]S_5-\dfrac13S_5[/tex], the difference reduces to
[tex]\dfrac23S_5=-\left(1-\dfrac1{3^5}\right)[/tex]
[tex]\implies S_5=\dfrac32\left(\dfrac1{3^5}-1\right)=\dfrac1{2\cdot3^4}-\dfrac32=\dfrac1{162}-\dfrac32=-\dfrac{121}{81}[/tex]
so the answer is E.
