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

2 Answers
Jun 21, 2016

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

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

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

"Step4: Put it all together"Step4: Put it all together

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

Let t_iti be any term in the whole series
Let the last term be t_ntn
Let the 'seed value' be aa
Let rr be the geometric ratio
Note that as we are only trying to determine nn the switch between positive and negative is of no consequence.

=> t_i=t_1" "->" "ar^0= |+4|=4ti=t1 ar0=|+4|=4 thus a=4a=4
=>t_i=t_2" "->" "ar^1=|-8|=8ti=t2 ar1=|8|=8

The general term expression is: ar^(i-1)ari1

From this r=(ar^1)/(ar^0) = 8/4=2r=ar1ar0=84=2

Also the last term is:

t_n=ar^(n-1) ->4(2^(n-1))=|-2048|tn=arn14(2n1)=|2048|

=>2^(n-1)=2048/4=5122n1=20484=512

Taking log_10log10 ( does not matter if you use log_eloge)

(n-1)log(2)=log(512) -> n-1=9(n1)log(2)=log(512)n1=9

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

Jun 21, 2016

Solution sheet page 2 of 2

The sum is -1364

Explanation:

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

Let the sum be ss 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)