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Proving the Sum of Cubes by Induction

This lesson demonstrates how to use the principle of mathematical induction to prove the elegant identity for the sum of the first n cubes, building the logic from a base case to a general inductive step.

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The Sum of Cubes Problem
∑i=1ni3=[n(n+1)2]2\sum_{i=1}^{n} i^{3} = \left[\frac{n(n+1)}{2}\right]^{2}
Have you ever noticed that the sum of the first n cubes is actually the square of the sum of the first n integers? Let's explore this beautiful identity using the powerful method of induction.
Step-by-step solver
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(a) Base Case

Verify the statement for n = 1.

13=[1(2)/2]2=12=11^3 = [1(2)/2]^2 = 1^2 = 1
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(b) Inductive Hypothesis

Assume the formula holds for an integer k.

∑i=1ki3=[k(k+1)/2]2\sum_{i=1}^{k} i^3 = [k(k+1)/2]^2
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(c) Inductive Step Addition

Add the (k+1)^3 term to the hypothesis sum.

∑i=1k+1i3=[k(k+1)/2]2+(k+1)3\sum_{i=1}^{k+1} i^3 = [k(k+1)/2]^2 + (k+1)^3
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(d) Algebraic Simplification

Factor out (k+1)^2 and simplify the quadratic expression.

(k+1)2[k2/4+(k+1)]=(k+1)2[(k2+4k+4)/4]=[(k+1)(k+2)/2]2(k+1)^2 [k^2/4 + (k+1)] = (k+1)^2 [(k^2+4k+4)/4] = [(k+1)(k+2)/2]^2
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(e) Conclusion

The result matches the form [n(n+1)/2]^2 for n=k+1, completing the proof.

P(k)  ⟹  P(k+1)P(k) \implies P(k+1)

Original question

Prove by mathematical induction that 1^3 + 2^3 + ... + n^3 = [n(n+1)/2]^2 for all positive integers n.

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