 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ia/paper-4/1c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Let

T_{n} = n (n + 1) /2

, the

n

\sum_{r = 1}^{n} r = T_{n}

. We prove the cube identity by mathematical induction. For

n = 1

both sides equal one. If

\sum_{r = 1}^{n} r^{3} = T_{n}^{2}

, then

T_{n + 1}^{2} - T_{n}^{2} = \frac{( n + 1 ) ^{2}}{4} [(n + 2)^{2} - n^{2}] = (n + 1)^{3} .

Adding

(n + 1)^{3}

establishes the next case. Thus

r = 1 \sum n r^{3} = [\frac{n ( n + 1 )}{2}]^{2} = (r = 1 \sum n r)^{2} .

Equivalently, the same displayed difference telescopes from

T_{0} = 0

, providing a direct proof without induction.

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