# How do you find the sum of the finite geometric sequence of Sigma 2^(n-1) from n=1 to 9?

Feb 14, 2018

${\sum}_{n = 1}^{9} {2}^{n - 1} = 511$

#### Explanation:

In general:

$1 + x + {x}^{2} + \ldots + {x}^{n - 1} = \frac{{x}^{n} - 1}{x - 1}$

so:

${\sum}_{n = 1}^{9} {2}^{n - 1} = {\sum}_{n = 0}^{8} {2}^{n} = \frac{{2}^{9} - 1}{2 - 1} = 511$

Feb 14, 2018

$\setminus$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {\sum}_{n = 1}^{9} \setminus {2}^{n - 1} \setminus = 511.$

#### Explanation:

$\setminus$

$\text{Recall the result (formula) for the sum of a finite geometric}$
$\text{series:}$

 \quad sum_{k = 0}^n \ r^n \ = \ { r^{ n + 1 } - 1 } / { r - 1 }; \qquad \qquad \qquad "where" \ r \ "is the common ratio".

$\text{The sum you have fits exactly this, with some small}$
$\text{adjustments.}$
$\text{Notice that the fundamental formula above starts with the}$
$\text{index variable" \ [ k ] \ \ "at "0," and your sum starts with the}$
$\text{index variable" \ [ n ] \ \ "at "1.}$

$\text{So let's start with your sum, and make the small adjustments}$
$\text{so we can use the fundamental formula above:}$

$\setminus q \quad \setminus q \quad \setminus q \quad {\sum}_{n = 1}^{9} \setminus {2}^{n - 1} \setminus = \setminus \left[{2}^{0 - 1} \setminus + \setminus {\sum}_{n = 1}^{9} \setminus {2}^{n - 1}\right] \setminus - \setminus {2}^{0 - 1} .$

$\text{Continuing, absorb the" \ \ 2^{ 0 - 1 } \ \ "term into the larger sum, since:}$
${2}^{0 - 1} = \setminus {2}^{n - 1} , \setminus \text{with} \setminus n = 0 :$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus \left[{\sum}_{n = 0}^{9} \setminus {2}^{n - 1}\right] \setminus - \setminus {2}^{- 1}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus \left[{\sum}_{n = 0}^{9} \setminus {2}^{- 1} \setminus \cdot {2}^{n}\right] \setminus - \setminus {2}^{- 1} .$

$\text{Continuing, pull out the" \ 2^{ - 1 } \ "factor inside the last sum:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus \left({2}^{- 1} \setminus \cdot \left[{\sum}_{n = 0}^{9} \setminus {2}^{n}\right]\right) \setminus - \setminus {2}^{- 1} .$

$\text{Now pull out the" \ 2^{ - 1 } \ "factor from the two terms here:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus {2}^{- 1} \setminus \cdot \left(\left[{\sum}_{n = 0}^{9} \setminus {2}^{n}\right] \setminus - \setminus 1\right) .$

$\text{Thus we have now:}$

$\setminus q \quad \setminus q \quad \setminus q \quad {\sum}_{n = 1}^{9} \setminus {2}^{n - 1} \setminus = \setminus {2}^{- 1} \setminus \cdot \left(\left[{\sum}_{n = 0}^{9} \setminus {2}^{n}\right] \setminus - \setminus 1\right) .$

$\text{Now the sum inside the brackets fits exactly the}$
$\text{fundamental formula above -- using:" \ \ r = 2, n = 9,"in that formula. So, continuing:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus {2}^{- 1} \setminus \cdot \left(\left[\frac{{2}^{9 + 1} - 1}{2 - 1}\right] \setminus - \setminus 1\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus {2}^{- 1} \setminus \cdot \left(\left[\frac{{2}^{10} - 1}{1}\right] \setminus - \setminus 1\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus {2}^{- 1} \setminus \cdot \left(\left({2}^{10} - 1\right) \setminus - \setminus 1\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus {2}^{- 1} \setminus \cdot \left({2}^{10} - 1 \setminus - \setminus 1\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus {2}^{- 1} \setminus \cdot \left({2}^{10} - 2\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus \left({2}^{- 1} \setminus \cdot {2}^{10}\right) - \left({2}^{- 1} \setminus \cdot 2\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus {2}^{10 - 1} - {2}^{1 - 1}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus {2}^{9} - {2}^{0}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus 512 - 1$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus 511.$

$\text{This our answer.}$

$\setminus$

$\text{Summarizing:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus {\sum}_{n = 1}^{9} \setminus {2}^{n - 1} \setminus = 511.$

$\setminus$

$\text{Note: There are other methods for doing this, using different}$
$\text{techniques, for example, one called changing the index of}$
$\text{summation. This is a good technique, and may be easier in}$
$\text{this case. The techniques displayed here illustrate alternatives}$
$\text{that can be powerful, and are quite frequently very convenient.}$