004 Sample Final A, Problem 13

Compute $\displaystyle {\sum _{n=1}^{6}4\left({\frac {1}{2}}\right)^{n}}$

Foundations
What is the formula for the sum of the first n terms of a geometric sequence?
Answer:
The sum of the first n terms of a geometric sequence is $S_{n}={\frac {A_{1}(1-r^{n})}{1-r}}$
where $r$ is the common ratio and $A_{1}$ is the first term of the geometric sequence.

Solution:

Step 1:
The common ratio for this geometric sequence is $r={\frac {1}{2}}$.
The first term of the geometric sequence is $4{\frac {1}{2}}=2$.

Step 2:
Using the above formula, we have
$\displaystyle {\sum _{n=1}^{6}4\left({\frac {1}{2}}\right)^{n}}=S_{6}={\frac {2(1-({\frac {1}{2}})^{6})}{(1-{\frac {1}{2}})}}$

Step 3:
If we simplify, we get
$\displaystyle {\sum _{n=1}^{6}4\left({\frac {1}{2}}\right)^{n}}={\frac {2(1-{\frac {1}{64}})}{\frac {1}{2}}}=4{\frac {63}{64}}={\frac {63}{16}}$.

Final Answer:
${\frac {63}{16}}$

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