allerySum of the CUBES of the first $n$ natural numbers, $1^3 + 2^3 + \ldots + n^3$, equals
A$n(n+1)/2$
B$n(n+1)(2n+1)/6$
C$n^4/4$
D$[n(n+1)/2]^2$
Answer & Solution
Correct answer: D. $[n(n+1)/2]^2$
1. Your syllabus: $\sum_{k=1}^{n} k^3 = \left(\dfrac{n(n+1)}{2}\right)^2$.
2. STRIKING IDENTITY: the sum of the first $n$ cubes equals the SQUARE of the sum of the first $n$ integers.
3. Example: $n = 3$. Sum of cubes: $1 + 8 + 27 = 36$. Square of sum of first 3 integers: $(1+2+3)^2 = 36$ ✓.
4. $n = 4$: cubes $= 1 + 8 + 27 + 64 = 100$. $(1+2+3+4)^2 = 10^2 = 100$ ✓.
5. Option A is linear sum. Option B is sum of squares. Option C is approximate but wrong.
_Source: syllabus Mathematics, Ch 8, §8.5 (Sum of cubes — Eq. 8.29), p. 13._
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