What are the first five partial sums of the geometric series (5/8)*3^(n-1)?

Any help is appreciated!

~Caylie

Originally Posted by Caylie
What are the first five partial sums of the geometric series (5/8)*3^(n-1)?

Any help is appreciated!

~Caylie
$\displaystyle \sum_{n=1}^5 \left(\frac{5}{8} \cdot 3^{n-1}\right)=\frac{5}{8} \cdot \displaystyle \sum_{n=1}^5 3^{n-1}=\frac{5}{8} \cdot \left(3^0+3^1+3^2+3^3+3^4\right)=\frac{605}{8}$

Originally Posted by princeps
$\displaystyle \sum_{n=1}^5 \left(\frac{5}{8} \cdot 3^{n-1}\right)=\frac{5}{8} \cdot \displaystyle \sum_{n=1}^5 3^{n-1}=\frac{5}{8} \cdot \left(3^0+3^1+3^2+3^3+3^4\right)=\frac{605}{8}$
This is the fifth partial sum. The first four are formed similarly.