# 004 Sample Final A, Problem 13

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

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

Solution:

Step 1:
The common ratio for this geometric sequence is ${\displaystyle r={\frac {1}{2}}}$.
The first term of the geometric sequence is ${\displaystyle 4{\frac {1}{2}}=2}$.
Step 2:
Using the above formula, we have
${\displaystyle \displaystyle {\sum _{n=1}^{6}4\left({\frac {1}{2}}\right)^{n}}=S_{6}={\frac {2(1-({\frac {1}{2}})^{6})}{(1-{\frac {1}{2}})}}}$
Step 3:
If we simplify, we get
${\displaystyle \displaystyle {\sum _{n=1}^{6}4\left({\frac {1}{2}}\right)^{n}}={\frac {2(1-{\frac {1}{64}})}{\frac {1}{2}}}=4{\frac {63}{64}}={\frac {63}{16}}}$.
${\displaystyle {\frac {63}{16}}}$