Exercise 7.1 — Question 6

Principle of Mathematical Induction

To prove a statement $P(n)$ by Mathematical Induction, follow three steps:

1. Base Case: Verify $P(1)$ (or the initial value) is true.
2. Inductive Hypothesis: Assume $P(k)$ is true for some positive integer $k$.
3. Inductive Step: Using the hypothesis, prove $P(k+1)$ is true.
4. Conclusion: By the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

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Question 6

Prove by Mathematical Induction:

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

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Step 1: Base Case ($n=1$)

LHS: $1^3=1$

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

Since LHS $=$ RHS $=1$, the statement is true 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:

$\underset{\text{by hypothesis}}{\underbrace{1^3+2^3+\cdots +k^3}}+(k+1{)}^3=\frac{k^2(k+1{)}^2}{4}+(k+1{)}^3$

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

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

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

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

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

This equals the RHS for $n=k+1$. ✓

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Conclusion

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

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