What are the first five partial sums of the geometric series (5/8)*3^(n-1)? Any help is appreciated! ~Caylie
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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 $\displaystyle \displaystyle \sum_{n=1}^5 \left(\frac{5}{8} \cdot 3^{n-1}\right)=\frac{5}{8} \cdot \displaystyle \sum_{n=1}^5 3^{n-1}=\frac{5}{8} \cdot \left(3^0+3^1+3^2+3^3+3^4\right)=\frac{605}{8}$
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.
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