Exercise 2: Dynamic Array Doubling for n=32 — Possible Solution
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GIVEN
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32 appends, starting from capacity 1, doubling capacity at each
resize (per this chapter's own method).

STEP 1: THE RESIZE EVENTS AND THEIR COPY COSTS
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Capacity 1 -> 2:  copy 1 element
Capacity 2 -> 4:  copy 2 elements
Capacity 4 -> 8:  copy 4 elements
Capacity 8 -> 16: copy 8 elements
Capacity 16 -> 32: copy 16 elements

Total copy cost: 1 + 2 + 4 + 8 + 16 = 31

STEP 2: THE ELEMENT-PLACEMENT COST
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Every one of the 32 appends also does the actual O(1) work of placing
the new element itself: 32 placements.

STEP 3: TOTAL OPERATIONS
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Total = 31 (copying) + 32 (placements) = 63

STEP 4: THE AMORTIZED COST PER APPEND
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63 / 32 ~= 1.97 operations per append

This is consistent with this chapter's own finding that the amortized
cost per append hovers around a small constant (roughly 2) regardless
of n - matching the chapter's own n=16 (1.94), n=100 (2.27), and
n=1000 (2.02) results closely.

WHY THIS WORKS AS AN ANSWER
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Each resize event's own copy cost is listed individually rather than
computed as a single lump sum, the placement cost is added separately,
and the final amortized figure is compared directly against this
chapter's own previously reported results at other values of n to
confirm the O(1) amortized pattern holds here too.