#3, -6, 12, -24, ...#

You can rewrite this as:

#(-1)^n 3*(1, 2, 4, 8, ...)#

If we focus on #1, 2, 4, 8...#

#a_(n+1) = 2a_n#

with #a_0 = 1#.

This can be written out as a nice sum:

#sum_(n=0)^(N) 2^n = 2^0 + 2^1 + 2^2 + 2^3 + ...#

#= 1 + 2 + 4 + 8 + ...#

Thus, now we can recombine everything to get:

#color(green)(3sum_(n=0)^(N) (-1)^n 2^n)#

where #N# is some arbitrary stop point.

The formula for summing this is:

#r = -2#

(since #(-1)^n * 2^n = (-2)^n#)

#a = 3#

#n = 10#

#=> 3((1-(-2)^10))/(1+2)# #-> (-2)^10 = 2^10#

#= 3((1-2^(5*2))/3)#

#= cancel(3)(1-32^2)/cancel(3) = 1-1024 = color(blue)(-1023)#

To sum this up more easily than brute-forcing it, if you don't remember the formula, there's something rather creative you can do.

If you look at the odd-indexed terms (#-6, -24, -96, -384,# etc), you can see that they are all successively multiplied by #4#. You can see that the even terms (#3, 12, 48, 192,# etc) are also this way. Thus, it lines up like so:

#n = 0 " " " 2 " " " 4 " " " " 6#:

#" " 3 + 12 + 48 + 192 + ...#

#n = 1 " " " 3 " " " 5 " " " " 7#:

#-6 -24 -96 -384 - ...#

and you always subtract twice the value of the positive number. Thus, just add up the first five even-indexed numbers (indices #0, 2, 4, ...#) and switch the sign.

#= -(3 + 12 + 48 + 192 + 768)#

#= -3(4^n)#

with #n_0 = 0#

#= -3(4^n) = -3[4^0 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5]#

#= -3(1 + 4 + 16 + 64 + 256)#

#= -3(341) = color(blue)(-1023)#

(You didn't have to write down as many terms here.)

Or, try adding up the original to check:

#= 3 - 6 + 12 - 24 + 48 - 96 + 192 - 384 + 768 - 1536)#

#= -3 - 12 - 48 - 192 - 768#

#= -15 - 240 - 768#

# = -255 - 768#

#= -223 - 800 = color(blue)(-1023)#

(Except now you had to write out 10 terms to add up.)