Toolkit 3

Sum of cubes

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

Proof

Let

T_k = 1 + 2 + \cdots + k = \frac{k(k+1)}{2}

denote the $k$ -th triangular number, where $T_0 = 0$ .

Then

\begin{aligned} T_k^2 - T_{k-1}^2 &= \left(\frac{k(k+1)}{2}\right)^2 - \left(\frac{k(k-1)}{2}\right)^2 \\ &= \frac{k^2}{4}\left((k+1)^2 - (k-1)^2\right) \\ &= \frac{k^2}{4}(4k) \\ &= k^3. \end{aligned}

Therefore,

k^3 = T_k^2 - T_{k-1}^2.

Writing this identity for $k = 1, 2, \ldots, n$ and adding, we obtain

\begin{aligned} 1^3 + 2^3 + \cdots + n^3 &= (T_1^2 - T_0^2) + (T_2^2 - T_1^2) + \cdots + (T_n^2 - T_{n-1}^2) \\ &= T_n^2. \end{aligned}

Since

T_n = \frac{n(n+1)}{2},

it follows that

1^3 + 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2. \quad\square