Source is Time Institute practice test

Mechmeera wrote:S1 is a square of side 256 cm. By joining its midpoints another square S2 was formed. By joining the midpoints of S2 another square S3 was formed. What is the excess of the total area of S1, S3, S5, S7, S9 over the total area of S2, S4, S6, S8, S10? (in sq cm)

128 (487)

128 (393)

128 (341)

64 (413)

64 (457)

This problem seems WAY too convoluted for the GMAT.
That said, it does provide a useful take-away:
When faced with a complex situation, test an EASY CASE.

Let each side of S� = 4.
The following figure is implied:

Every triangle in the figure above is a 45-45-90 triangle.
In a 45-45-90 triangle, the sides are in the following ratio:
s - s - s√2.

Resulting areas:
S� = 4² = 16.
S₂ = (2√2)² = 8.
S₃ = 2² = 4.
S₄ = (√2)² = 2.

Thus:
S� + S₃ = 16+4 = 20.
Sâ‚‚ + Sâ‚„ = 8+2 = 10.
Difference between the two sums = 20-10 = 10.

Note the values in red.
The difference between the two sums is equal to the sum of the EVEN-NUMBERED squares.
Thus, the problem above requires that we determine the following sum:
S₂ + S₄ + S₆ + S₈ + S�₀.

Actual areas:
The figure above illustrates that each even-numbered square is equal to 1/2 the area of the preceding odd-numbered square.
Since the actual area of S� = (256)(256), the actual area of S₂ = (128)(256).

The figure also illustrates that the area of each even-numbered square is to equal to 1/4 the area of the preceding even-numbered square.
Thus:
S₄ = (¼)(S₂) = (¼)(128)(256) = 32*256.
S₆ = (¼)(S₄) = (¼)(32)(256) = 8*256.
S₈ = (¼)(S₆) = (¼)(8)(256)= 2*256.
S�₀ = (¼)(S₈) = (¼)(2)(256)= (½)(256).

Resulting sum:
(128)(256) + 32*256 + 8*256 + 2*256 + (½)(256)

= (256)(128 + 32 + 8 + 2 + ½)

= (256)(170.5)

= (256/2)(2*170.5)

= (128)(341).

The correct answer is C.

Alternate approach:
Since the area of each even-numbered square is to equal to 1/4 the area of the preceding even-numbered square, S₂, S₄, S₆, S₈, and S�₀ constitute a GEOMETRIC SEQUENCE in which the multiplier between successive terms is ¼.

For any geometric sequence of n terms with a multiplier of r, the SUM of the first n terms is equal to the following:
(first term)[ (1 - r^n)/(1 - r) ].

Since S₂ + S₄ + S₆ + S₈ + S�₀ is composed of n=5 terms with a multiplier of r=1/4, we get:

(first term)[ (1 - r^n)/(1 - r) ] =

= (256)(128) * [ (1 - (¼)�)/(1 - ¼) ]

= (2�)(2�) * [ (1 - (½)¹�)/(¾) ]

= (2�)(2�) * [ (2¹� - 1)/2¹�) ] * (4/3)

= (2�)(2�) * [ 1023/2¹� ] * (2²/3)

= 2� * 341

= 128 * 341.