2020HSCShort answer

Q14 (c)

4 marks

Question

Prove by mathematical induction that, for $n\geq2$ ,
$\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{n^2}<\frac{n-1}{n}.$

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Solution

Criteria	Marks
Provides correct solution	4
Correctly proves the inductive step<br>OR<br>Verifies initial case, $n=2$ , AND assumes true for $k$ and attempts to verify true for $k+1$	3
Assumes true for $k$ and attempts to verify true for $k+1$	2
Verifies initial case, $n=2$	1

Sample answer:

Let $P(n)$ be the given proposition.

$P(2)$ is true since

\frac1{2^2}<\frac12

Let $k$ be an integer for which $P(k)$ is true. Thus we assume that

\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{k^2}<\frac{k-1}{k}

Consider

\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{k^2}+\frac1{(k+1)^2}

<\frac1{(k+1)^2}+\frac{k-1}{k}

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

<\frac{(k+1)+(k+1)^2(k-1)}{(k+1)^2k}

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

\therefore P(k+1)\text{ is true.}

\therefore P(n)\text{ true for all integers }n\ge2\text{ by induction.}

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Working MathematicallyProof

Further Proof by Mathematical InductionUse mathematical induction for series, divisibility, inequalities, algebra, calculus, probability, geometry and recursive formulae.

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