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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