8A F11 Q14

Question: Compute $\displaystyle {\sum _{n=1}^{\infty }5\left({\frac {3}{5}}\right)^{n}}$

Foundations
1) What type of series is this?
2) Which formula, on the back page of the exam, is relevant to this question?
3) In the formula there are some placeholder variables. What is the value of each placeholder?
Answer:
1) This series is geometric. The giveaway is there is a number raised to the nth power.
2) The desired formula is $S_{\infty }={\frac {a_{1}}{1-r}}$
3) $a_{1}$ is the first term in the series, which is $5{\frac {3}{5}}=3$. The value for r is the ratio between consecutive terms, which is ${\frac {3}{5}}$

Solution:

Step 1:
We start by identifying this series as a geometric series, and the desired formula for the sum being $S_{\infty }={\frac {a_{1}}{1-r}}$.

Step 2:
Since $a_{1}$ is the first term in the series, $a_{1}=5{\frac {3}{5}}=3$. The value for r is the ratio between consecutive terms, which is ${\frac {3}{5}}$. Plugging everything in we have $S_{\infty }={\frac {3}{1-{\frac {3}{5}}}}={\frac {3}{\frac {2}{5}}}={\frac {15}{2}}$

Final Answer:
${\frac {15}{2}}$

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