I have one question about a geometric series...please help. :)

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

Any help is appreciated!

~Caylie :)

Re: I have one question about a geometric series...please help. :)

Quote:

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 :)

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

Re: I have one question about a geometric series...please help. :)

Quote:

Originally Posted by

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

This is the fifth partial sum. The first four are formed similarly.