we have been asked to find the sum of the series
[tex] \sum _{n=1}^5\left(\frac{1}{3}\right)^{n-1} [/tex]
As we know that a geometric series has a constant ratio "r" and it is defined as
[tex] r=\frac{a_{n+1}}{a_n}=\frac{\left(\frac{1}{3}\right)^{\left(n+1\right)-1}}{\left(\frac{1}{3}\right)^{n-1}}=\frac{1}{3} [/tex]
The first term of the series is [tex] a_1=\left(\frac{1}{3}\right)^{1-1}=1 [/tex]
Geometric series sum formula is
[tex] S_n=a_1\frac{1-r^n}{1-r} [/tex]
Plugin the values we get
[tex] S_5=1\cdot \frac{1-\left(\frac{1}{3}\right)^5}{1-\frac{1}{3}} [/tex]
On simplification we get
[tex] S_5=\frac{121}{81} [/tex]
Hence the sum of the given series is [tex] \frac{121}{81} [/tex]
