Induction (1 Viewer)

AbstractBlade · Mar 24, 2021

If anyone can solve question 40. (ii) it would be appreciated

AbstractBlade · Mar 24, 2021

If anyone can solve question 39 it would be appreciated

AbstractBlade · Mar 24, 2021

If anyone can solve question 38 it would be appreciated. I suck at induction.

CM_Tutor · Mar 24, 2021

Question 38

Theorem: For all integers positive integers $n$

$\sum_{r=1}^n r^3 = 1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}$

Proof: By induction on $n$

A... Test $n = 1$

$\text{LHS } = \sum_{r=1}^1 r^3 = 1^3 = 1$

$\text{RHS } = \frac{n^2(n+1)^2}{4} = \frac{1^2 \times 2^2}{4} = \frac{4}{4} = 1 = \text{ LHS}$

So, the result is true for $n = 1$

B... Let $k$ be a value of $n$ for which the result is true. That is,

$\sum_{r=1}^k r^3 = 1^3 + 2^3 + 3^3 + ... + k^3 = \frac{k^2(k+1)^2}{4} \qquad \qquad \text{. . . . . . . . . (**)}$

We must now prove the result for $n = k+1$ . That is, we must prove that

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

$\begin{align*} \text{LHS } &= \sum_{r=1}^{k+1} r^3 \\ &= 1^3 + 2^3 + 3^3 + ... + (k+1)^3 \\ &= \left(1^3 + 2^3 + 3^3 + ... + k^3\right) + (k+1)^3 \\ &= \frac{k^2(k+1)^2}{4} + (k+1)^3 \qquad \text{using the induction hypothesis (**)} \\ &= \frac{(k+1)^2}{4}\left[k^2 + 4(k+1)\right] \\ &= \frac{(k+1)^2(k^2 + 4k+4)}{4} \\ &= \frac{(k+1)^2(k+2)^2}{4} \\ &= \text{ RHS} \end{align*}$

So if the result is true for $n=k$ , then it must also be true for $n=k+1$ .

C... It follows from A and B by the process of mathematical induction that the result is true for all positive integers $n$ .

AbstractBlade · Mar 24, 2021

CM_Tutor said:
Question 38

Theorem: For all integers positive integers $n$

$\sum_{r=1}^n r^3 = 1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}$

Proof: By induction on $n$

A... Test $n = 1$

$\text{LHS } = \sum_{r=1}^1 r^3 = 1^3 = 1$

$\text{RHS } = \frac{n^2(n+1)^2}{4} = \frac{1^2 \times 2^2}{4} = \frac{4}{4} = 1 = \text{ LHS}$

So, the result is true for $n = 1$

B... Let $k$ be a value of $n$ for which the result is true. That is,

$\sum_{r=1}^k r^3 = 1^3 + 2^3 + 3^3 + ... + k^3 = \frac{k^2(k+1)^2}{4} \qquad \qquad \text{. . . . . . . . . (**)}$

We must now prove the result for $n = k+1$ . That is, we must prove that

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

$\begin{align*} \text{LHS } &= \sum_{r=1}^{k+1} r^3 \\ &= 1^3 + 2^3 + 3^3 + ... + (k+1)^3 \\ &= \left(1^3 + 2^3 + 3^3 + ... + k^3\right) + (k+1)^3 \\ &= \frac{k^2(k+1)^2}{4} + (k+1)^3 \qquad \text{using the induction hypothesis (**)} \\ &= \frac{(k+1)^2}{4}\left[k^2 + 4(k+1)\right] \\ &= \frac{(k+1)^2(k^2 + 4k+4)}{4} \\ &= \frac{(k+1)^2(k+2)^2}{4} \\ &= \text{ RHS} \end{align*}$

So if the result is true for $n=k$ , then it must also be true for $n=k+1$ .

C... It follows from A and B by the process of mathematical induction that the result is true for all positive integers $n$ .

Thank you, this helps a lot.

CM_Tutor · Mar 24, 2021

Question 39

Theorem: For all integers $n \geqslant 2$

$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... + \frac{1}{n^2} < \frac{n-1}{n}$

Proof: By induction on $n$

A... Test $n = 2$

$\text{LHS } = \frac{1}{2^2} = \frac{1}{4}$

$\text{RHS } = \frac{n-1}{n} = \frac{1}{2} > \frac{1}{4} = \text{ LHS}$

So, the result is true for $n = 2$

B... Let $k$ be a value of $n$ for which the result is true. That is,

$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... + \frac{1}{k^2} < \frac{k-1}{k} \qquad \qquad \text{. . . . . . . . . (**)}$

We must now prove the result for $n = k+1$ . That is, we must prove that

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

$\begin{align*} \text{LHS } &= \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... + \frac{1}{(k+1)^2} \\ &= \left(\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... + \frac{1}{k^2}\right) + \frac{1}{(k+1)^2} \\ &< \frac{k-1}{k} + \frac{1}{(k+1)^2} \qquad \text{using the induction hypothesis (**)} \\ &= \frac{1}{(k+1)^2}\left[\frac{(k-1)(k+1)^2}{k} + 1\right] \\ &= \frac{(k-1)(k+1)^2+k}{k(k+1)^2} \\ &= \frac{(k - 1)(k^2+2k+1) + k}{k(k+1)^2} \\ &= \frac{k^3 + k^2 - 1}{k(k+1)^2} \\ &= \frac{k^2(k+1) - 1}{k(k+1)^2} \\ &= \frac{k}{k+1} - \frac{1}{k(k+1)^2} \\ &< \frac{k}{k+ 1} \qquad \text{as $\frac{1}{k(k+1)^2} > 0$}\\ &= \text{ RHS} \end{align*}$

So if the result is true for $n=k$ , then it must also be true for $n=k+1$ .

C... It follows from A and B by the process of mathematical induction that the result is true for all positive integers $n \geqslant 2$ .

CM_Tutor · Mar 25, 2021

Question 40

Theorem: cot x - cot 2x = cosec 2x

Proof:

$\begin{align*} \text{LHS } &= \cot{x} - \cot{2x} \\ &= \frac{1}{\tan{x}} - \frac{1}{\tan{2x}} \\ &= \frac{\cos{x}}{\sin{x}} - \frac{\cos{2x}}{\sin{2x}} \\ &= \frac{\cos{x} \times 2\cos{x}}{\sin{x} \times 2\cos{x}} - \frac{2\cos^2{x} - 1}{\sin{2x}} \qquad \qquad \text{as $\cos{2x} = 2\cos^2{x} - 1$} \\ &= \frac{2\cos^2{x}}{\sin{2x}} + \frac{1 - 2\cos^2{x}}{\sin{2x}} \qquad \qquad \text{as $2\sin{x}\cos{x} = \sin{2x}$} \\ &= \frac{1}{\sin{2x}} \\ &= \text{cosec}\ x \\ &= \text{ RHS} \end{align*}$

Note that this result can be generalised:

$\begin{align*} \cot{kx} - \cot{2kx} &= \frac{1}{\tan{kx}} - \frac{1}{\tan{2kx}} \\ &= \frac{\cos{kx}}{\sin{kx}} - \frac{\cos{2kx}}{\sin{2kx}} \\ &= \frac{\cos{kx} \times 2\cos{kx}}{\sin{kx} \times 2\cos{kx}} - \frac{2\cos^2{kx} - 1}{\sin{2kx}} \qquad \qquad \text{using $\cos{2\alpha} = 2\cos^2{\alpha} - 1$ with $\alpha = kx$} \\ &= \frac{2\cos^2{kx}}{\sin{2kx}} + \frac{1 - 2\cos^2{kx}}{\sin{2kx}} \qquad \qquad \text{as $2\sin{kx}\cos{kx} = \sin{2kx}$} \\ &= \frac{1}{\sin{2kx}} \\ \text{So,} \quad \cot{kx} - \cot{2kx} &= \text{cosec}\ 2kx \qquad \qquad \text{. . . . . . . . . (*)} \end{align*}$

The theorem that we are given can then be solved used telescoping series, without the need for induction, though an induction proof is also valid:

Theorem: For all positive integers $n$

$\sum_{r=1}^n \text{cosec}\ 2^rx = \cot{x} - \cot{2^nx}$

Proof without induction:

$\begin{align*} \text{LHS } &= \sum_{r=1}^n \text{cosec}\ 2^rx \\ &= \text{cosec}\ 2x + \text{cosec}\ 4x + \text{cosec}\ 8x + ... + \text{cosec}\ 2^nx \\ &= \left(\cot{x} - \cot{2x}\right) + \left(\cot{2x} - \cot{4x}\right) + \left(\cot{4x} - \cot{8x}\right) + ... + \left(\cot{2^{n-1}x} - \cot{2^nx}\right) \quad \text{using (*) with (sequentially) $k = 1, 2, 4, ..., 2^{n-1}$} \\ &= \cot{x} + \left(-\cot{2x} + \cot{2x}\right) + \left(-\cot{4x} + \cot{4x}\right) + \left(-\cot{8x} + \cot{8x}\right) + ... + \left(-\cot{2^{n-1}x} + \cot{2^{n-1}x}\right) - \cot{2^nx} \\ &= \cot{x} + 0 + 0 + 0 + ... + 0 - \cot{2^nx} \\ &= \cot{x} - \cot{2^nx} \\ &= \text{ RHS} \end{align*}$

Proof by induction (on $n$ ):

A... Test $n = 1$

$\text{LHS } = \sum_{r=1}^1 \text{cosec}\ 2^1x = \text{cosec}\ 2x = \cot{x} - \cot{2x} \qquad \qquad \text{as was shown in part (a)}$

$\text{RHS } = \cot{x} - \cot{2^1x} = \cot{x} - \cot{2x} = \text{ LHS}$

So, the result is true for $n = 1$

B... Let $k$ be a value of $n$ for which the result is true. That is,

$\begin{align*} \sum_{r=1}^k \text{cosec}\ 2^rx &= \text{cosec}\ 2x + \text{cosec}\ 4x + \text{cosec}\ 8x + ... + \text{cosec}\ 2^kx \\ &= \cot{x} - \cot{2^kx} \qquad \qquad \text{. . . . . . . . . (**)} \end{align*}$

We must now prove the result for $n = k+1$ . That is, we must prove that

$\begin{align*} \sum_{r=1}^{k+1} \text{cosec}\ 2^rx &= \text{cosec}\ 2x + \text{cosec}\ 4x + \text{cosec}\ 8x + ... + \text{cosec}\ 2^{k+1}x \\ &= \cot{x} - \cot{2^{k+1}x} \end{align*}$

$\begin{align*} \text{LHS } &= \sum_{r=1}^{k+1} \text{cosec}\ 2^rx \\ &= \text{cosec}\ 2x + \text{cosec}\ 4x + \text{cosec}\ 8x + ... + \text{cosec}\ 2^{k+1}x \\ &= \left(\text{cosec}\ 2x + \text{cosec}\ 4x + \text{cosec}\ 8x + ... + \text{cosec}\ 2^kx\right) + \text{cosec}\ 2^{k+1}x \\ &= \cot{x} - \cot{2^kx} + \text{cosec}\ 2^{k+1}x \qquad \text{using the induction hypothesis (**)} \\ &= \cot{x} - \cot{2^kx} + \cot{\frac{2^{k+1}x}{2}} - \cot{2^{k+1}x} \qquad \qquad \text{using (*)} \\ &= \cot{x} - \cot{2^kx} + \cot{2^kx} - \cot{2^{k+1}x} \\ &= \cot{x} - \cot{2^{k+1}x} \\ &= \text{ RHS} \end{align*}$

So if the result is true for $n=k$ , then it must also be true for $n=k+1$ .

C... It follows from A and B by the process of mathematical induction that the result is true for all positive integers $n$ .

AbstractBlade · Mar 27, 2021

CM_Tutor said:
Question 40

Theorem: cot x - cot 2x = cosec 2x

Proof:

$\begin{align*} \text{LHS } &= \cot{x} - \cot{2x} \\ &= \frac{1}{\tan{x}} - \frac{1}{\tan{2x}} \\ &= \frac{\cos{x}}{\sin{x}} - \frac{\cos{2x}}{\sin{2x}} \\ &= \frac{\cos{x} \times 2\cos{x}}{\sin{x} \times 2\cos{x}} - \frac{2\cos^2{x} - 1}{\sin{2x}} \qquad \qquad \text{as $\cos{2x} = 2\cos^2{x} - 1$} \\ &= \frac{2\cos^2{x}}{\sin{2x}} + \frac{1 - 2\cos^2{x}}{\sin{2x}} \qquad \qquad \text{as $2\sin{x}\cos{x} = \sin{2x}$} \\ &= \frac{1}{\sin{2x}} \\ &= \text{cosec}\ x \\ &= \text{ RHS} \end{align*}$

Note that this result can be generalised:

$\begin{align*} \cot{kx} - \cot{2kx} &= \frac{1}{\tan{kx}} - \frac{1}{\tan{2kx}} \\ &= \frac{\cos{kx}}{\sin{kx}} - \frac{\cos{2kx}}{\sin{2kx}} \\ &= \frac{\cos{kx} \times 2\cos{kx}}{\sin{kx} \times 2\cos{kx}} - \frac{2\cos^2{kx} - 1}{\sin{2kx}} \qquad \qquad \text{using $\cos{2\alpha} = 2\cos^2{\alpha} - 1$ with $\alpha = kx$} \\ &= \frac{2\cos^2{kx}}{\sin{2kx}} + \frac{1 - 2\cos^2{kx}}{\sin{2kx}} \qquad \qquad \text{as $2\sin{kx}\cos{kx} = \sin{2kx}$} \\ &= \frac{1}{\sin{2kx}} \\ \text{So,} \quad \cot{kx} - \cot{2kx} &= \text{cosec}\ 2kx \qquad \qquad \text{. . . . . . . . . (*)} \end{align*}$

The theorem that we are given can then be solved used telescoping series, without the need for induction, though an induction proof is also valid:

Theorem: For all positive integers $n$

$\sum_{r=1}^n \text{cosec}\ 2^rx = \cot{x} - \cot{2^nx}$

Proof without induction:

$\begin{align*} \text{LHS } &= \sum_{r=1}^n \text{cosec}\ 2^rx \\ &= \text{cosec}\ 2x + \text{cosec}\ 4x + \text{cosec}\ 8x + ... + \text{cosec}\ 2^nx \\ &= \left(\cot{x} - \cot{2x}\right) + \left(\cot{2x} - \cot{4x}\right) + \left(\cot{4x} - \cot{8x}\right) + ... + \left(\cot{2^{n-1}x} - \cot{2^nx}\right) \quad \text{using (*) with (sequentially) $k = 1, 2, 4, ..., 2^{n-1}$} \\ &= \cot{x} + \left(-\cot{2x} + \cot{2x}\right) + \left(-\cot{4x} + \cot{4x}\right) + \left(-\cot{8x} + \cot{8x}\right) + ... + \left(-\cot{2^{n-1}x} + \cot{2^{n-1}x}\right) - \cot{2^nx} \\ &= \cot{x} + 0 + 0 + 0 + ... + 0 - \cot{2^nx} \\ &= \cot{x} - \cot{2^nx} \\ &= \text{ RHS} \end{align*}$

Proof by induction (on $n$ ):

A... Test $n = 1$

$\text{LHS } = \sum_{r=1}^1 \text{cosec}\ 2^1x = \text{cosec}\ 2x = \cot{x} - \cot{2x} \qquad \qquad \text{as was shown in part (a)}$

$\text{RHS } = \cot{x} - \cot{2^1x} = \cot{x} - \cot{2x} = \text{ LHS}$

So, the result is true for $n = 1$

B... Let $k$ be a value of $n$ for which the result is true. That is,

$\begin{align*} \sum_{r=1}^k \text{cosec}\ 2^rx &= \text{cosec}\ 2x + \text{cosec}\ 4x + \text{cosec}\ 8x + ... + \text{cosec}\ 2^kx \\ &= \cot{x} - \cot{2^kx} \qquad \qquad \text{. . . . . . . . . (**)} \end{align*}$

We must now prove the result for $n = k+1$ . That is, we must prove that

$\begin{align*} \sum_{r=1}^{k+1} \text{cosec}\ 2^rx &= \text{cosec}\ 2x + \text{cosec}\ 4x + \text{cosec}\ 8x + ... + \text{cosec}\ 2^{k+1}x \\ &= \cot{x} - \cot{2^{k+1}x} \end{align*}$

$\begin{align*} \text{LHS } &= \sum_{r=1}^{k+1} \text{cosec}\ 2^rx \\ &= \text{cosec}\ 2x + \text{cosec}\ 4x + \text{cosec}\ 8x + ... + \text{cosec}\ 2^{k+1}x \\ &= \left(\text{cosec}\ 2x + \text{cosec}\ 4x + \text{cosec}\ 8x + ... + \text{cosec}\ 2^kx\right) + \text{cosec}\ 2^{k+1}x \\ &= \cot{x} - \cot{2^kx} + \text{cosec}\ 2^{k+1}x \qquad \text{using the induction hypothesis (**)} \\ &= \cot{x} - \cot{2^kx} + \cot{\frac{2^{k+1}x}{2}} - \cot{2^{k+1}x} \qquad \qquad \text{using (*)} \\ &= \cot{x} - \cot{2^kx} + \cot{2^kx} - \cot{2^{k+1}x} \\ &= \cot{x} - \cot{2^{k+1}x} \\ &= \text{ RHS} \end{align*}$

So if the result is true for $n=k$ , then it must also be true for $n=k+1$ .

C... It follows from A and B by the process of mathematical induction that the result is true for all positive integers $n$ .

If you were able to help with another question it would be greatly appreciated. It's question 31.

Thank you

Qeru · Mar 28, 2021

AbstractBlade said:
If you were able to help with another question it would be greatly appreciated. It's question 31.

Thank you

Not CM but:
(i):
$\text{RHS}=(x-1)\left(\frac{x^n-1}{x-1}-n\right) \quad \text{Using the GP sum formula}$
$\text{RHS}=x^n-1-n(x-1)=\text{LHS}$

(ii): Let: $f(x)=(x-1)(1+x+x^2+...+x^{n-1}-n)$ . Since x is a positive integer, the minimum value of x must be 1. The minimum value of f(x) is: $f(1)=0$ as f(x) is clearly an increasing function. Therefore: $f(x) \geq 0$ which implies: $x^n-1-n(x-1) \geq 0$ from part i and rearranging gives the desired result.

(iii)

Substitute $x=\frac{a}{b}$ into the result found in ii.

Therefore:
$\left(\frac{a}{b}\right)^n\geq 1+n\left (\frac{a}{b}-1\right)$

$a^nb^{-n}b \geq b\left(1+n\left (\frac{a}{b}-1\right)\right) \quad \text{multiply both sides by b, note that b is greater than zero}$

$a^nb^{1-n} \geq na+(1-n)b \quad \text{after expanding and rearranging}$

AbstractBlade · Mar 28, 2021

Qeru said:
Not CM but:
(i):
$\text{RHS}=(x-1)\left(\frac{x^n-1}{x-1}-n\right) \quad \text{Using the GP sum formula}$
$\text{RHS}=x^n-1-n(x-1)=\text{LHS}$

(ii): Let: $f(x)=(x-1)(1+x+x^2+...+x^{n-1}-n)$ . Since x is a positive integer, the minimum value of x must be 1. The minimum value of f(x) is: $f(1)=0$ as f(x) is clearly an increasing function. Therefore: $f(x) \geq 0$ which implies: $x^n-1-n(x-1) \geq 0$ from part i and rearranging gives the desired result.

(iii)

Substitute $x=\frac{a}{b}$ into the result found in ii.

Therefore:
$\left(\frac{a}{b}\right)^n\geq 1+n\left (\frac{a}{b}-1\right)$

$a^nb^{-n}b \geq b\left(1+n\left (\frac{a}{b}-1\right)\right) \quad \text{multiply both sides by b, note that b is greater than zero}$

$a^nb^{1-n} \geq na+(1-n)b \quad \text{after expanding and rearranging}$

thank you
absolute legend

CM_Tutor · Mar 28, 2021

Qeru said:
Not CM but:
(i):
$\text{RHS}=(x-1)\left(\frac{x^n-1}{x-1}-n\right) \quad \text{Using the GP sum formula}$
$\text{RHS}=x^n-1-n(x-1)=\text{LHS}$

(ii): Let: $f(x)=(x-1)(1+x+x^2+...+x^{n-1}-n)$ . Since x is a positive integer, the minimum value of x must be 1. The minimum value of f(x) is: $f(1)=0$ as f(x) is clearly an increasing function. Therefore: $f(x) \geq 0$ which implies: $x^n-1-n(x-1) \geq 0$ from part i and rearranging gives the desired result.

(iii)

Substitute $x=\frac{a}{b}$ into the result found in ii.

Therefore:
$\left(\frac{a}{b}\right)^n\geq 1+n\left (\frac{a}{b}-1\right)$

$a^nb^{-n}b \geq b\left(1+n\left (\frac{a}{b}-1\right)\right) \quad \text{multiply both sides by b, note that b is greater than zero}$

$a^nb^{1-n} \geq na+(1-n)b \quad \text{after expanding and rearranging}$

I certainly agree with @Qeru in general. However, the question declares that $x$ is a positive real, not a positive integer... it is $n$ that must be an integer. Thus, Qeru's proof based on $x$ being an integer and being at least 1 is flawed.

Further, I would be careful in part (ii) in using a phrase like "clearly an increasing function." Examiners generally prefer the obvious to be explained / justified without asserting something is clear. Further, since an increasing function can be defined as:

$f(x)$ is increasing on $[a, b]$ if $f'(x) > 0\ \forall x \in \[a, b]$

then the lack of any derivative in the answer could become a problem.

The approach I would use in this question is:

I can define

$f(x) = x^n - 1 - n(x - 1)$

and

$g(x) = (x - 1)\left(1 + x + x^2 + ... + x^{n-1} - n\right)$ ,

in both cases where $n\in\mathbb{Z}^+$ and $x\in\mathbb{R}^+$ ,

and know from part (i) that $f(x)=g(x)$ under these conditions.

I seek to demonstrate that $f(x) \geqslant 0$ and will do this by instead establishing that $g(x)$ has this property:

Case 1: $x = 1$

$g(1) = (1 - 1)\left(1 + 1 + 1^2 + ... + 1^{n-1} - n\right) = 0$

Case 2: $x \neq 1, x > 0$

I can change the " $n$ " term in $g(x)$ into $n$ distinct ones (ie $n = 1 + 1 + 1 + ... + 1$ ) and then rearrange to get

$g(x) = (x-1)\left[(1-1)+(x-1)+(x^2-1)+...+(x^{n-1}-1)\right]$

Case 2(a): $x > 1$

$\begin{align*} g(x) &= (\text{+ve term})\left[0 + (\text{+ve term}) + (\text{+ve term}) + ... + (\text{+ve term})\right] \\ &= (\text{+ve term}) \times (\text{+ve term}) \\ &> 0 \end{align*}$

Case 2(b): $0 < x < 1$

$\begin{align*} g(x) &= (\text{-ve term})\left[0 + (\text{-ve term}) + (\text{-ve term}) + ... + (\text{-ve term})\right] \\ &= (\text{-ve term}) \times (\text{-ve term}) \\ &> 0 \end{align*}$

So, case 1 gives $g(x)=0$ when $x=1$ and case 2 gives $g(x)>0$ if $x\in(0, 1)\cup(1, +\infty)$ and thus $g(x)\geqslant 0$ for $x\in\mathbb{R}^+$ .

And, then, since $f(x)=g(x)$ throughout this domain, $f(x) \geqslant 0\ \forall x\in\mathbb{R}^+$ .

$\begin{align*} f(x) = x^n - 1 - n(x - 1) &\geqslant 0 \\ x^n &\geqslant 1 + n(x - 1) \end{align*}$

Paradoxica · Mar 29, 2021

CM_Tutor said:
I certainly agree with @Qeru in general. However, the question declares that $x$ is a positive real, not a positive integer... it is $n$ that must be an integer. Thus, Qeru's proof based on $x$ being an integer and being at least 1 is flawed.

Further, I would be careful in part (ii) in using a phrase like "clearly an increasing function." Examiners generally prefer the obvious to be explained / justified without asserting something is clear. Further, since an increasing function can be defined as:

$f(x)$ is increasing on $[a, b]$ if $f'(x) > 0\ \forall x \in \[a, b]$

then the lack of any derivative in the answer could become a problem.

Personally I don't see a problem with this, but you would have to prove that a for f a strictly increasing function, f applied to a strictly increasing sequence is a strictly increasing sequence. However, I feel like this lemma has been stated without proof in past HSC questions, so I don't know.

CM_Tutor · Mar 29, 2021

Paradoxica said:
Personally I don't see a problem with this, but you would have to prove that a for f a strictly increasing function, f applied to a strictly increasing sequence is a strictly increasing sequence. However, I feel like this lemma has been stated without proof in past HSC questions, so I don't know.

If $x$ is a positive integer then @Qeru's proof is fine and $f(x)$ does strictly increase from a minimum of $f(1)=0$ provided $n>1$ . If $n = 1$ then $f(x) = 0$ throughout its domain and the final result is trivial.

However, $x\in\mathbb{R}^+$ and $f(x)$ does not strictly increase on $(0, +\infty)$ for integers $n \geqslant 2$ .

Consider $f(x)$ when $n=2$ : $f(x)= (x-1)(1+x-2)=(x-1)^2$

Now $f(x)$ is a parabola and has a minimum of $f(1)=0$ but is decreasing on $(0, 1)$ .

Qeru's proof does not run into such cases because there is no integer below 1 in the domain $(0, +\infty)$ but, as the question defines $x$ as real, the fact that

$f\left(\frac{1}{2}\right)>f(1)=0\ \forall n\geqslant 2$

(to take one example) makes the assertion that $f(x)$ strictly increases false.

Edited to add: The difference between $x$ being a positive real as against a positive integer is essential to the last part of the question. The substitution $x=\frac{a}{b}$ where $a, b \in\mathbb{R}^+$ is basically pointless if $x$ is required to be an integer because it amounts to requiring that $b=1$ . Any other case will have $b$ as a factor of $a$ and can thus be reduced to an equivalent $b=1$ case.

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