Sum of Sequence of Cubes/Proof by Recursion

Theorem

$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$

Proof

From Closed Form for Triangular Numbers‎:

$(1): \quad \ds \map A n := \sum_{i \mathop = 1}^n i = \frac{n \paren {n + 1} } 2$

From Sum of Sequence of Squares:

$(2): \quad \ds \map B n := \sum_{i \mathop = 1}^n i^2 = \frac{n \paren {n + 1} \paren {2 n + 1} } 6$

Let $\ds \map S n = \sum_{i \mathop = 1}^n i^3$.

Then:

\(\ds \map S n\)

\(=\)

\(\ds n^3 + \paren {n - 1}^3 + \paren {n - 2}^3 + \cdots + 2^3 + 1^3\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^{n \mathop - 1} \paren {n - k}^3\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^{n \mathop - 1} \paren {n^3 - 3 n^2 k + 3 n k^2 - k^3}\)

\(\ds \)

\(=\)

\(\ds n^4 - 3 n^2 \cdot \map A {n - 1} + 3 n \cdot \map B {n - 1} - \map S {n - 1}\)

\(\text {(3)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \map S n\)

\(=\)

\(\ds n^4 - 3 n^2 \cdot \frac {n \paren {n - 1} } 2 + 3n \cdot \frac{n \paren {n - 1} \paren {2 n - 1} } 6 - \map S {n - 1}\)

substituting from $(1)$ and $(2)$

\(\text {(4)}: \quad\)

\(\ds \map S n\)

\(=\)

\(\ds n^3 + \map S {n - 1}\)

recursive definition

\(\ds \leadsto \ \ \)

\(\ds 2 \map S n\)

\(=\)

\(\ds n^4 + n^3 - 3 n^2 \cdot \frac {n \paren {n - 1} } 2 + 3 n \cdot \frac {n \paren {n - 1} \paren {2 n - 1} } 6\)

adding $(3)$ and $(4)$

\(\ds \leadsto \ \ \)

\(\ds \)

\(=\)

\(\ds \frac {n^2 \paren {n + 1}^2} 2\)

simplification

\(\ds \leadsto \ \ \)

\(\ds \map S n\)

\(=\)

\(\ds \frac {n^2 \paren {n + 1}^2} 4\)

$\blacksquare$