# What is the sum of the first 6 terms of this geometric sequence? 1,8,64,512

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Tony B Share
Feb 10, 2018

$37449$

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#### Explanation:

$\textcolor{b l u e}{\text{Understanding how the sequence works}}$

Let the term count be $i$
Let the ${i}^{\text{th}}$ term be ${a}_{i}$

Then we have:

$i = 1 \to {a}_{1} = 1$
$i = 2 \to {a}_{2} = 8$
$i = 3 \to {a}_{3} = 64$
$i = 4 \to {a}_{4} = 512$

The rate of increase is far to great for an arithmetic progression so it is a geometric sequence as stated in the question.

Lets have a play!

From our multiplication tables notice that $8 \times 8 = 64$ so this could be a link.

Lets try ${8}^{3} = 512$ which works. So we have something involving $\times 8$

At a guess lets investigate:

${a}_{1} = 1 \textcolor{g r e e n}{\leftarrow \text{ I will come back to this}}$
${a}_{2} = 8 = 1 \times 8$
${a}_{3} = 64 = 1 \times 8 \times 8$
${a}_{4} = 512 = 1 \times 8 \times 8 \times 8$

So by observation, what is happening is that we have for any term where $i > 1 \textcolor{w h i t e}{\text{dd}}$ ${a}_{i} = 1 \times {8}^{i - 1}$

Given that ${a}_{1} = 1$
${a}_{i} = {a}_{2} = 8 = {a}_{1} \times {8}^{2 - 1} = {8}^{1}$
${a}_{i} = {a}_{3} = 64 = {a}_{1} \times {8}^{3 - 1} = {8}^{2}$
${a}_{i} = {a}_{4} = 512 = {a}_{1} \times {8}^{4 - 1} = {8}^{3}$

How does ${a}_{1}$ fit this process. Note that ${8}^{0} = 1$ so for
${a}_{i} = {a}_{1} = 1 \times {8}^{1 - 1} \textcolor{w h i t e}{\text{dd")=color(white)("dd")1xx8^0color(white)("dd")=color(white)("dd}} 1 \times 1 = 1$

So it works for ${a}_{1}$ as well thus we are now are working with $i > 0$

Thus the general rule is ${a}_{i} = 1 \times {8}^{i - 1} = {8}^{i - 1}$
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$\textcolor{b l u e}{\text{Determine the sum to any term } {a}_{n}}$

Set $s = 1 + {8}^{1} + {8}^{2} + {8}^{3} + \ldots {8}^{n - 1} \text{ } \ldots \ldots . E q u a t i o n \left(1\right)$

Multiply $s$ by 8 giving:

$8 s = {8}^{1} + {8}^{2} + {8}^{3} + \ldots {8}^{n - 1} + {8}^{n} \text{ } \ldots \ldots E q u a t i o n \left(2\right)$

Subtract $E q n \left(2\right) \text{ from } E q n \left(1\right)$

$8 s - s = {8}^{n} - 1$

Factor out $s$

$s \left(8 - 1\right) = {8}^{n} - 1$

$s = \frac{{8}^{n} - 1}{7}$

$\textcolor{b r o w n}{\text{So summing to the "6^("th")" term we have:}}$

$s = \frac{{8}^{6} - 1}{7}$

$\textcolor{w h i t e}{\text{dddddddddd}} \textcolor{b l u e}{\underline{\overline{| \textcolor{w h i t e}{\frac{2}{2}} s = 37449 \textcolor{w h i t e}{\frac{2}{2}} |}}}$
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$\textcolor{b l u e}{\text{Check}}$

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Jim G. Share
Feb 9, 2018

${S}_{6} = 37449$

#### Explanation:

$\text{using the sum to n terms for a geometric sequence}$

•color(white)(x)S_n=(a(1-r^n))/(1-r)=(a(r^n-1))/(r-1)

$\text{where a is the first term and r the common ratio}$

$r = {a}_{2} / {a}_{1} = {a}_{3} / {a}_{2} = \ldots \ldots = {a}_{n} / {a}_{n - 1}$

$\Rightarrow r = \frac{8}{1} = \frac{64}{8} = \frac{512}{64} = 8 \text{ and } a = 1$

$\Rightarrow {S}_{6} = \frac{1 \left({8}^{6} - 1\right)}{7} = 37449$

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Feb 9, 2018

${\sum}_{k = 1}^{6} \left(1\right) {8}^{k - 1} = 37449$

#### Explanation:

From the reference Geometric Progression we obtain the equation:

${a}_{n} = a {r}^{n - 1}$

We can find the value of a; given that ${a}_{1} = 1$

${a}_{1} = a {r}^{1 - 1}$

$1 = a {r}^{0}$

$a = 1$

We can find the value of r; given that ${a}_{2} = 8$ and knowing that $a = 1$

${a}_{2} = \left(1\right) {r}^{2 - 1}$

$r = 8$

The same reference gives us the equation:

${\sum}_{k = 1}^{n} a {r}^{k - 1} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}$

We know that $a = 1$ and $r = 8$ and we want the first 6 terms which means that $n = 6$:

${\sum}_{k = 1}^{6} \left(1\right) {8}^{k - 1} = \frac{1 \left(1 - {8}^{6}\right)}{1 - 8}$

${\sum}_{k = 1}^{6} \left(1\right) {8}^{k - 1} = 37449$

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