Question

Question #85f5f

Answer 1

`sum_(k=1)^6 (1)8^(k-1)= 37449`

Explanation:

From the reference Geometric Progression we obtain the equation:

`a_n = ar^(n-1)`

We can find the value of a; given that `a_1 = 1`

`a_1 = ar^(1-1)`

`1 =ar^0`

`a = 1`

We can find the value of r; given that `a_2 = 8` and knowing that `a = 1`

`a_2 = (1)r^(2-1)`

`r = 8`

The same reference gives us the equation:

`sum_(k=1)^n ar^(k-1)= (a(1-r^n))/(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 (1)8^(k-1)= (1(1-8^6))/(1-8)`

`sum_(k=1)^6 (1)8^(k-1)= 37449`

Answer 2

`S_(6)=37449`

Explanation:

`"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)`

`"where a is the first term and r the common ratio"`

`r=a_2/a_1=a_3/a_2= ...... =a_n/a_(n-1)`

`rArrr=8/1=64/8=512/64=8" and "a=1`

`rArrS_6=(1(8^6-1))/7=37449`

Answer 3

`37449`

ALL background information given

Explanation:

`color(blue)("Understanding how the sequence works")`

Let the term count be `i`
Let the `i^("th")` term be `a_i`

Then we have:

`i=1->a_1=1`
`i=2->a_2=8`
`i=3->a_3=64`
`i=4->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 `8xx8=64` so this could be a link.

Lets try `8^3=512` which works. So we have something involving `xx8`

At a guess lets investigate:

`a_1=1color(green)(larr" I will come back to this")`
`a_2=8=1xx8`
`a_3=64=1xx8xx8`
`a_4=512=1xx8xx8xx8`

So by observation, what is happening is that we have for any term where `i>1color(white)("dd")` `a_i=1xx8^(i-1)`

Given that `a_1=1`
`a_i=a_2=8=a_1xx8^(2-1)=8^1`
`a_i=a_3=64=a_1xx8^(3-1)=8^2`
`a_i=a_4=512=a_1xx8^(4-1)=8^3`

How does `a_1` fit this process. Note that `8^0=1` so for
`a_i=a_1=1xx8^(1-1)color(white)("dd")=color(white)("dd")1xx8^0color(white)("dd")=color(white)("dd")1xx1=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=1xx8^(i-1) = 8^(i-1)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the sum to any term " a_n)`

Set `s=1+8^1+8^2+8^3+...8^(n-1)" ".......Equation(1)`

Multiply `s` by 8 giving:

`8s=8^1+8^2+8^3+...8^(n-1)+8^n" "......Equation(2)`

Subtract `Eqn(2)" from "Eqn(1)`

`8s-s=8^n-1`

Factor out `s`

`s(8-1)=8^n-1`

`s=(8^n-1)/(7)`

`color(brown)("So summing to the "6^("th")" term we have:")`

`s=(8^6-1)/(7)`

`color(white)("dddddddddd")color(blue)( ul(bar(|color(white)(2/2)s=37449color(white)(2/2)|)))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Check")`