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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