The sum of the cubes of the first \( n \) natural numbers is given by: \( \displaystyle S_n = \left( \frac{n(n + 1)}{2} \right)^2 \). The formula shows that the sum of the first \( n \) cubes is equal to the square of the sum of the first \( n \) natural numbers. This identity is sometimes also called Nicomachus’s Theorem. In this post, we will learn how to prove it.
Proof:
We will prove it using mathematical induction.
Base Case \( ( n = 1 ) \):
\( \displaystyle S_1 = 1^3 = 1 = \left( \frac{1(1 + 1)}{2} \right)^2 = 1^2 = 1 \)
The formula holds for \( n = 1 \).
Inductive Step:
Assume the formula holds for \( n = k \), i.e.,
\( \displaystyle S_k = 1^3 + 2^3 + \dots + k^3 = \left( \frac{k(k + 1)}{2} \right)^2 \)
Prove for \( n = k + 1 \):
\( \begin{aligned} &S_{k+1} = S_k + (k + 1)^3 \\ &= \left( \frac{k(k + 1)}{2} \right)^2 + (k + 1)^3 \\ &= \frac{k^2(k + 1)^2}{4} + (k + 1)^3 \\ &= \frac{k^2(k + 1)^2 + 4(k + 1)^3}{4} \\ &= \frac{(k + 1)^2 (k^2 + 4(k + 1))}{4} \\ &= \frac{(k + 1)^2 (k^2 + 4k + 4)}{4} \\ &= \frac{(k + 1)^2 (k + 2)^2}{4} \\ &= \left( \frac{(k + 1)(k + 2)}{2} \right)^2 \\ \end{aligned} \)Thus, the formula holds for \( n = k + 1 \).
By induction, the formula is true for all \( n \in \mathbb{N} \).
Please let me know in the comments if you find any errors in this proof.