Question

Find the sum of the finite geometric sequence

Answer 1

$\displaystyle\sum_{n=1}^5-\left(\frac13\right)^{n-1}=-\left(1+\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}\right)$

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

$\dfrac13S_5=-\left(\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}+\dfrac1{3^5}\right)$

so if we consider $S_5-\dfrac13S_5$, the difference reduces to

$\dfrac23S_5=-\left(1-\dfrac1{3^5}\right)$

$\implies S_5=\dfrac32\left(\dfrac1{3^5}-1\right)=\dfrac1{2\cdot3^4}-\dfrac32=\dfrac1{162}-\dfrac32=-\dfrac{121}{81}$

so the answer is E.