Find the geometric sum 4 + 12 + 36 + . . + 236,196. 1,062,880 4,194,303 354,292 118,096

The answer choices are sufficiently far apart that you can work this backward. The sum will be ...
  236,196*(1 + 1/3 + 1/9 + 1/27 + ...)
so a reasonable estimate can be given by an infinite series with a common ratio of 1/3. That sum is
  236,196*(1/(1 - 1/3)) = 236,196*(3/2)

Without doing any detailed calculation, you know the best answer choice is ...
  354,292


There are log(236196/4)/log(3) + 1 = 11 terms* in the series, so the sum will be found to be 4(3^11 -1)/(3-1) = 2*(3^11-1) = 354,292.

Using the above approach (working backward from the last term), the sum will be 236,196*(1-(1/3)^11)/(1-(1/3)) = 236,196*1.49999153246 = 354,292

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* If you just compute log(236196/4)/log(3) = 10 terms, then your sum comes out 118,096--a tempting choice. However, you must realize that the last term is larger than this, so this will not be the sum. (In fact, the sum is this value added to the last term.)