Exercise 7.1 — Question 5

Mathematical Induction

Mathematical Induction is a proof technique used to establish that a statement $P(n)$ is true for all positive integers $n$. It consists of three steps:

1. Base Case: Verify that $P(1)$ (or the initial value) is true.
2. Inductive Hypothesis: Assume $P(k)$ is true for some arbitrary positive integer $k$.
3. Inductive Step: Using the hypothesis, prove that $P(k+1)$ is also true.
4. Conclusion: Since $P(1)$ is true and $P(k)\Rightarrow P(k+1)$, by the Principle of Mathematical Induction, $P(n)$ is true for all $n\in \mathbb{N}$.

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Question 5 — Prove by Mathematical Induction

Statement: Prove that for all $n\in \mathbb{N}$: $1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}$

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

LHS: $1^2=1$

RHS: $\frac{1(1+1)(2\cdot 1+1)}{6}=\frac{1\cdot 2\cdot 3}{6}=\frac{6}{6}=1$

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: $1^2+2^2+3^2+\cdots +k^2=\frac{k(k+1)(2k+1)}{6}$

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

We need to show: $1^2+2^2+\cdots +k^2+(k+1{)}^2=\frac{(k+1)(k+2)(2k+3)}{6}$

Starting from the LHS: $\underset{\text{use hypothesis}}{\underbrace{1^2+2^2+\cdots +k^2}}+(k+1{)}^2$

$=\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2$

$=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1{)}^2}{6}$

$=\frac{(k+1)[k(2k+1)+6(k+1)]}{6}$

$=\frac{(k+1)[2k^2+k+6k+6]}{6}$

$=\frac{(k+1)(2k^2+7k+6)}{6}$

Factorise $2k^2+7k+6$: $2k^2+7k+6=(k+2)(2k+3)$

Therefore: $=\frac{(k+1)(k+2)(2k+3)}{6}$

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

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Conclusion

Since the statement holds for $n=1$ (base case), and assuming it holds for $n=k$ implies it holds for $n=k+1$ (inductive step), by the Principle of Mathematical Induction, the statement $1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}$ is true for all $n\in \mathbb{N}$.