Minton Analytics

Theorem

For any positive integer \(n\), the sum of the first \(n\) cubes is exactly equal to the square of the sum of the first \(n\) integers.

∑k=1nk3=(∑k=1nk)2=(n(n+1)2)2\sum_{k=1}^{n} k^3 = \left(\sum_{k=1}^{n} k\right)^2 = \left(\frac{n(n+1)}{2}\right)^2

Proof

We prove by induction that \(\displaystyle\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2\) for all \(n \geq 1\).

Base Case: Let \(n = 1\).

∑k=11k3=13=1and(1(1+1)2)2=(22)2=1\sum_{k=1}^{1} k^3 = 1^3 = 1 \quad \text{and} \quad \left(\frac{1(1+1)}{2}\right)^2 = \left(\frac{2}{2}\right)^2 = 1

The base case holds.

Inductive Step: We show the statement holds for \(n = m + 1\).

Inductive Hypothesis: Assume the statement holds for \(n = m\), i.e., assume

∑k=1mk3=(m(m+1)2)2\sum_{k=1}^{m} k^3 = \left(\frac{m(m+1)}{2}\right)^2
∑k=1m+1k3=∑k=1mk3+(m+1)3\sum_{k=1}^{m+1} k^3 = \sum_{k=1}^{m} k^3 + (m+1)^3

By the inductive hypothesis

=(m(m+1)2)2+(m+1)3= \left(\frac{m(m+1)}{2}\right)^2 + (m+1)^3
=(m(m+1)2)2+(m+1)(m+1)2= \left(\frac{m(m+1)}{2}\right)^2 + (m+1)(m+1)^2
=(m(m+1)2)2+m(m+1)2+(m+1)2= \left(\frac{m(m+1)}{2}\right)^2 + m(m+1)^2+(m+1)^2
=(m(m+1)2)2+2(m(m+1)2)(m+1)+(m+1)2= \left(\frac{m(m+1)}{2}\right)^2 + 2\left(\frac{m(m+1)}{2}\right)(m+1)+(m+1)^2

Recognizing \(a^2+2ab+b^2\)

=(m(m+1)2+(m+1))2= \left(\frac{m(m+1)}{2}+(m+1)\right)^2
=((m+1)(m+2)2)2= \left(\frac{(m+1)(m+2)}{2}\right)^2
=(∑k=1m+1k)2= \left(\sum_{k=1}^{m+1} k\right)^2

Therefore, by induction, the statement holds for all \(n \geq 1\). ■

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