7.1 Q-8

Exercise 7.1 — Question 8

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 ($n=1$)

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

We must prove the statement holds for $n=k+1$, i.e., 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 $\frac{(k+1{)}^2}{4}$:

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

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

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

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

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Conclusion

Since the base case holds for $n=1$, and the truth of the statement for $n=k$ implies its truth for $n=k+1$, by the Principle of Mathematical Induction, the statement is true for all positive integers $n$.

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

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Key Structure of a Mathematical Induction Proof