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

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)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#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)#