# How do you find the sum of a finite geometric sequence from n = 1 to n = 8, using the expression −2(3)^n − 1?

Dec 19, 2016

I could just use the formula, plug in the values and get the answer. Not going to do that!

The sum is $\text{ } - 19688$

#### Explanation:

So that we can link this to the generic formula bit let the geometric ratio of 3 be $r$

Then we have:
${\sum}_{i = 1 \to n} \left[- 2 {r}^{i} - 1\right]$

${\sum}_{i = 1 \to n} - 2 {r}^{i} - {\sum}_{i = 1 \to n} 1$

but ${\sum}_{i = 1 \to n} 1 = n$ giving

${\sum}_{i = 1 \to n} \left[- 2 {r}^{i}\right] - n$

$- \left\{2 {\sum}_{i = 1 \to n} \left[{r}^{i}\right] + n\right\} \ldots \ldots \ldots \ldots E x p r e s s i o n 1$
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Set $\text{ } t = {\sum}_{i = 1 \to n} \left[{r}^{i}\right]$

$\text{ } t = {r}^{1} + {r}^{2} + {r}^{3} + \ldots + {r}^{n}$

$\text{ } r t = {r}^{2} + {r}^{3} + \ldots + {r}^{n} + {r}^{n + 1}$

$r t - t = {r}^{n + 1} - r$

$t = \frac{r \left({r}^{n} - 1\right)}{r - 1}$
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Thus by substitution in $E x p r e s s i o n \left(1\right)$ we have

$- \left\{2 {\sum}_{i = 1 \to n} \left[{r}^{i}\right] + n\right\} \text{ } \to - \left\{\textcolor{w h i t e}{\frac{2}{2}} 2 t + n\right\}$

$\left(- 1\right) \times \left\{2 \times \frac{3 \left({3}^{8} - 1\right)}{3 - 1} + 8\right\}$

$\left(- 1\right) \times \left\{{\cancel{2}}^{1} \times \frac{3 \left(6560\right)}{{\cancel{2}}^{1}} + 8\right\}$

$= - 19688$
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Check: 