Question

How do you find the sum of the terms of geometric 4-8+16-32+.....-2048?

Answer 1

Solution sheet page 1 of 2

Explanation:

We have 3 condition in this question and at the moment I can only think of 1 way of solving this.

`color(blue)("Method plan")`
I am going to solve this by splitting the series into a positive only series and into a negative only series. The combining their sums will result in the sum of the whole. I will start by building an equation for the sum.

`"Step1: Consider the whole sequence to determine how many terms"` `" there are."`

`"Step2: Consider the sum of all the positive value and develop the"` `" equation "`

`"Step3: Consider the sum of all the negative value and develop the"` `" equation "`

`"Step4: Put it all together"`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step1")`

Let `t_i` be any term in the whole series
Let the last term be `t_n`
Let the 'seed value' be `a`
Let `r` be the geometric ratio
Note that as we are only trying to determine `n` the switch between positive and negative is of no consequence.

`=> t_i=t_1" "->" "ar^0= |+4|=4` thus `a=4`
`=>t_i=t_2" "->" "ar^1=|-8|=8`

The general term expression is: `ar^(i-1)`

From this `r=(ar^1)/(ar^0) = 8/4=2`

Also the last term is:

`t_n=ar^(n-1) ->4(2^(n-1))=|-2048|`

`=>2^(n-1)=2048/4=512`

Taking `log_10` ( does not matter if you use `log_e`)

`(n-1)log(2)=log(512) -> n-1=9`

`color(blue)( n=10)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Answer 2

Solution sheet page 2 of 2

The sum is -1364

Explanation:

`color(Magenta)("Building the equation for the sum of the series")`

Let the sum be `s` then:

`s= ar^0+ar^1+ar^2+...+ar^(n-1)`..............Equation (1)

Multiply `s` by `r` giving:
`sr=ar^1+ar^2+ar^3...+ar^(n-1)+ar^n`...........Equetion (2)

Equation(2)-Equation(1) gives:

`sr-s" "=" "ar^n-a`

`color(magenta)(s=(a(r^n-1))/(r-1))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 2: Solving the positive part only")`
`color(green)("The whole series is count 10 so the positive will be count 5")`

`t_1->ar^0=4" " ->" " a=4`

`t_2=>ar^1=16`

`r=(ar^1)/(ar^0) = 16/4=4`

`color(brown)(=>s^+=(a(r^n-1))/(r-1))color(blue)( " "->" " (4(4^5-1) )/(4-1) = 1364`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 3: Solving the negative part only")`
`color(green)("The whole series is count 10 so the positive will be count 5")`

This will be the same as step 2 but in this case
`a=-8`
`r`
`color(brown)(s^(-)=(a(r^n-1))/(r-1))color(blue)(" "->" "-[(8(4^5-1))/(4-1) ]=-2728`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 4: Putting it all together")`

`color(magenta)(s=s^+ + s^-" "=" "1364-2728" "=" "-1364)`