Exercise 7.1 — Question 3

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 arbitrary positive integer $k$.
3. Inductive Step: Using the hypothesis, prove $P(k+1)$ is 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 positive integers $n$.

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Question 3: Prove by Mathematical Induction

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

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:

$\underset{\text{by hypothesis}}{\underbrace{1^2+2^2+\cdots +k^2}}+(k+1{)}^2=\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2$

Factor out $(k+1)$:

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

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

Expand the numerator:

$k(2k+1)+6(k+1)=2k^2+k+6k+6=2k^2+7k+6$

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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Step 4 — 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 positive integers $n$.