Detailed Answer

Let's consider five consecutive integers \( n-2 \), \( n-1 \), \( n \), \( n+1 \), and \( n+2 \).

The sum of these integers is:

\[(n-2) + (n-1) + n + (n+1) + (n+2) = 5n\]

So, the mean is:

\[\frac{5n}{5} = n\]

Since \( n \) is an integer, the mean is always an integer.

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Detailed Answer

To show that \((3n-2)^3 + (3n+2)^3 = 54n^3 + 72n\), we start by expanding each cube:

\[ (3n-2)^3 = 27n^3 - 54n^2 + 36n - 8 \]

\[ (3n+2)^3 = 27n^3 + 54n^2 + 36n + 8 \]

So, now we can add the two expressions:

\[ (3n-2)^3 + (3n+2)^3 = 54n^3 + 72n \]

Thus, we have shown that:

\[ (3n-2)^3 + (3n+2)^3 = 54n^3 + 72n \]

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Detailed Answer

a)

We start with the given: \((x+3)^2 = x^2 + 6x + 9\).

Then, we multiply by \((x+3)\):

\[ (x+3)^3 = (x^2 + 6x + 9)(x + 3) \]

Expand and combine like terms:

\[ (x+3)^3 = x^3 + 9x^2 + 27x + 27 \]

b)

i. \(x = 3\)

Direct calculation: \((3+3)^3 = 6^3 = 216\)

Substituting into the expanded form: \[ 3^3 + 9 * 3^2 + 27 * 3 + 27 = 27 + 81 + 81 + 27 = 216 \]

ii. \(x = -2\)

Direct calculation: \((-2+3)^3 = 1^3 = 1\)

Substituting into the expanded form: \[ (-2)^3 + 9 \cdot (-2)^2 + 27 \cdot (-2) + 27 = -8 + 36 - 54 + 27 = 1 \]

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Detailed Answer

We can express any odd number in the form \(2k + 1\), where \(k\) is an integer. So, let:

\[a = 2m + 1\]

\[b = 2n + 1\]

Where \(m\) and \(n\) are any integers.

The sum of \(a\) and \(b\) is:

\[a + b = (2m + 1) + (2n + 1)\]

\[a + b = 2m + 2n + 2 = 2(m+n+1)\]

Since \(m + n + 1\) is an integer, it has to be even since it is multiplied by 2. Thus, \(a + b\) is an even number.

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Detailed Answer

a)

\[ \frac{16x^2 - 4}{4x^2 + 6x + 2} = \frac{4x - 2}{x + 1} \]

Numerator: \( 16x^2 - 4 = (4x - 2)(4x+2) \)

Denominator: \( 4x^2 + 6x + 2 = (4x+2)(x+1) \)

\[ \frac{16x^2 - 4}{4x^2 + 6x + 2} = \frac{(4x - 2)(4x+2)}{(4x+2)(x+1)} \]

So, we can cancel \( 4x+2 \) from numerator and denominator, which ultimately proves the initial expression:

\[ \frac{16x^2 - 4}{4x^2 + 6x + 2} = \frac{4x - 2}{x + 1} \]

b)

We know that the denominator can't be equal to 0, so:

From the LHS: \( (4x+2)(x+1) \neq 0 \)

From the RHS: \( (x+1) \neq 0 \)

\[ (4x+2)(x+1) \neq 0 \]

\[ (x+1) \neq 0 \]

\[ x \neq -1, \ x \neq -\frac{1}{2} \]

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Detailed Answer

Let's consider three consecutive integers \( n-1 \), \( n \), and \( n+1 \). Their squares are:

\[(n-1)^2, n^2, (n+1)^2\]

By expanding these, we get:

\[(n-1)^2 = n^2 - 2n + 1\]

\[n^2\]

\[(n+1)^2 = n^2 + 2n + 1\]

The mean of these squares is:

\[\frac{(n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1)}{3} = \frac{3n^2 + 2}{3}\]

\(\frac{3n^2 + 2}{3}\) is not an integer because \(3n^2 + 2\) is not divisible by 3.

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Detailed Answer

Since the are two consecutive even integers, we know that \( b = a + 2 \). So, the difference of their squares can be denoted as:

\[ b^2 - a^2 \]

\[ (a+2)^2 - a^2 = a^2 + 4a + 4 - a^2 = 4a + 4 = 4(a+1) \]

As it can be noticed, \( a + 1 \) is the value of the integer between them, since the first integer is \( a \) and the second one is \( a + 2 \). Hence, we have proved that the difference between the squares is equal to 4 times the integer between \( a \) and \( b \).

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Detailed Answer

To prove the sum of cubes of two consecutive odd integers is divisible by 2, let the integers be \(2n+1\) and \(2n+3\).

Their cubes are: \((2n+1)^3\) and \((2n+3)^3\). Let's now expand them:

\[(2n+1)^3 = 8n^3 + 12n^2 + 6n + 1\]

\[(2n+3)^3 = 8n^3 + 8n^3 + 36n^2 + 54n + 27\]

By summing them up we have:

\[(2n+1)^3 + (2n+3)^3 = 16n^3 + 48n^2 + 60n + 28\]

Factoring out 2:

\[2(8n^3 + 24n^2 + 30n + 14)\]

Hence, the sum is divisible by 2.

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Detailed Answer

To show that \( 3\sqrt{2} - 5 \) is irrational, we will use a proof by contradiction.

Suppose \( 3\sqrt{2} - 5 \) is rational. Then, there exist integers \( a \) and \( b \) (with \( b \neq 0 \)) such that:

\[ 3\sqrt{2} - 5 = \frac{a}{b} \]

Rearrange the equation to solve for \( \sqrt{2} \):

\[ 3\sqrt{2} = \frac{a}{b} + 5 \]

\[ 3\sqrt{2} = \frac{a + 5b}{b} \]

\[ \sqrt{2} = \frac{a + 5b}{3b} \]

The right-hand side of the equation, \( \frac{a + 5b}{3b} \), is a rational number because \( a \), \( b \), and \( 5b \) are integers, and \( 3b \neq 0 \). However, \( \sqrt{2} \) is known to be irrational. This leads to a contradiction because a rational number cannot equal an irrational number.

Since our assumption that \( 3\sqrt{2} - 5 \) is rational leads to a contradiction, \( 3\sqrt{2} - 5 \) must be irrational.

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Detailed Answer

To prove that \( n^3 + 2n \) is a multiple of 3 for any integer \( n \), we will use a direct proof.

Any integer \( n \) can be expressed in one of the following forms:

1. \( n = 3k \)
2. \( n = 3k + 1 \)
3. \( n = 3k + 2 \)

where \( k \) is an integer.

Case 1: \( n = 3k \)

\[ n^3 + 2n = (3k)^3 + 2(3k) = 27k^3 + 6k = 3(9k^3 + 2k) \]

This is clearly a multiple of 3.

Case 2: \( n = 3k + 1 \)

\[ n^3 + 2n = (3k + 1)^3 + 2(3k + 1) \]

Expand \( (3k + 1)^3 \):

\[ (3k + 1)^3 = 27k^3 + 27k^2 + 9k + 1 \]

Thus:

\[ n^3 + 2n = 27k^3 + 27k^2 + 9k + 1 + 6k + 2 = 27k^3 + 27k^2 + 15k + 3 = 3(9k^3 + 9k^2 + 5k + 1) \]

This is a multiple of 3.

Case 3: \( n = 3k + 2 \)

\[ n^3 + 2n = (3k + 2)^3 + 2(3k + 2) \]

Expand \( (3k + 2)^3 \):

\[ (3k + 2)^3 = 27k^3 + 54k^2 + 36k + 8 \]

Thus:

\[ n^3 + 2n = 27k^3 + 54k^2 + 36k + 8 + 6k + 4 = 27k^3 + 54k^2 + 42k + 12 = 3(9k^3 + 18k^2 + 14k + 4) \]

This is a multiple of 3.

In all cases, \( n^3 + 2n \) is a multiple of 3. Therefore, \( n^3 + 2n \) is a multiple of 3 for any integer \( n \).

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Detailed Answer

Prove true for \( n = 1 \):

\[ \sum_{r=1}^1 r^3 = 1^3 = 1 \]

The right-hand side of the formula is:

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

Thus, the formula holds for \( n = 1 \).

Assume that the formula holds for some positive integer \( k \), i.e.,

\[ \sum_{r=1}^k r^3 = \left(\frac{k(k+1)}{2}\right)^2 \]

We need to show that the formula holds for \( k + 1 \), i.e.,

\[ \sum_{r=1}^{k+1} r^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2 \]

Start with the left-hand side:

\[ \sum_{r=1}^{k+1} r^3 = \sum_{r=1}^k r^3 + (k+1)^3 \]

Using the inductive hypothesis:

\[ \sum_{r=1}^{k+1} r^3 = \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3 \]

Simplify the expression:

\[ = \frac{k^2(k+1)^2}{4} + (k+1)^3 \]

Factor out \( (k+1)^2 \):

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

Combine the terms inside the parentheses:

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

Factor the numerator:

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

Thus, the formula holds for \( k + 1 \).

By the principle of mathematical induction, the formula

\[ \sum_{r=1}^n r^3 = \left(\frac{n(n+1)}{2}\right)^2 \]

holds for all positive integers \( n \).

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