Exercise 7.1 — Question 31

Problem Statement

Prove by Mathematical Induction that for all positive integers $n$:

$1^3+2^3+3^3+\cdots +n^3=\frac{n^2(n+1{)}^2}{4}$

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Structure of a Proof by Mathematical Induction

Every proof by Mathematical Induction has three parts:

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Proof

Step 1 — Base Case ($n=1$)

Left-Hand Side (LHS): $1^3=1$

Right-Hand Side (RHS): $\frac{1^2(1+1{)}^2}{4}=\frac{1\cdot 4}{4}=1$

Since LHS $=$ RHS $=1$, the statement holds for $n=1$. ✓

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Step 2 — Inductive Hypothesis

Assume the statement is true for $n=k$, i.e., assume:

$1^3+2^3+3^3+\cdots +k^3=\frac{k^2(k+1{)}^2}{4}$

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Step 3 — Inductive Step (Prove for $n=k+1$)

We need to show:

$1^3+2^3+\cdots +k^3+(k+1{)}^3=\frac{(k+1{)}^2(k+2{)}^2}{4}$

Starting from the LHS:

$\underset{\text{use inductive hypothesis}}{\underbrace{1^3+2^3+\cdots +k^3}}+(k+1{)}^3$

$=\frac{k^2(k+1{)}^2}{4}+(k+1{)}^3$

Factor out $(k+1{)}^2$:

$=(k+1{)}^2\left[\frac{k^2}{4}+(k+1)\right]$

$=(k+1{)}^2\left[\frac{k^2+4(k+1)}{4}\right]$

$=(k+1{)}^2\cdot \frac{k^2+4k+4}{4}$

$=(k+1{)}^2\cdot \frac{(k+2{)}^2}{4}$

$=\frac{(k+1{)}^2(k+2{)}^2}{4}$

This is exactly the RHS for $n=k+1$. ✓

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Conclusion

Since the statement is true for $n=1$ (base case), and assuming it is true for $n=k$ implies it is true for $n=k+1$ (inductive step), by the Principle of Mathematical Induction, the statement is true for all positive integers $n$:

$\boxed{1^3+2^3+3^3+\cdots +n^3=\frac{n^2(n+1{)}^2}{4}}$

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Key Takeaway