Math AA SL

The sum of the integers is:

k + (k + 2) = 2k + 2 = 2(k + 1)

Therefore the sum of these integers is even.

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To express the product of the sum and the difference of these integers we can write:

(a + b)(a - b)

Working further and simplifying, we get:

(a + b)(a - b) = a^{2} + ab - ab - b^{2} = a^{2} -  b^{2}

Therefore the product of the sum and the difference of these integers is equal to the difference of their squares.

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The sum of the squares of the integers is:

n^{2} + (n + 1)^{2} = n^{2} + n^{2} + 2n + 1 =

= 2n^{2} + 2n + 1 = 2(n^{2} + n) + 1

Therefore the sum of these integers is odd.

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(a)

Finding the sum of the integers:

n + n + 1 + n + 2 + n + 3 = 4n + 6

Factorising we get 2(2n + 3), therefore the sum of these four integers is divisible by 2, so it is an even number.

(b)

Finding the sum of the squares:

n^{2} + (n + 1)^{2} + (n + 2)^{2} + (n + 3)^{2}

n^{2} + (n^{2} + 2n + 1) + (n^{2} + 4n + 4) + (n^{2} + 6n + 9)

4n^{2} + 12n + 14

4n^{2} + 12n + 12 + 2 = 4(n^{2} + 3n + 3) + 2

Since 4(n^{2} + 3n + 3) is divisible by 4, 4(n^{2} + 3n + 3) + 2 is not divisible by 4.

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Sum of squares:

n^{2} + (n+2)^{2} = n^{2} + n^{2} + 4n + 4

Collecting terms and factorising 2:

2(n^{2} + 2n + 2)

Therefore the sum of the squares is divisible by two, hence it is an even number.

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Let’s represent the integers as n and n + 1

For the difference of the squares of the integers we get: 

(n + 1)^{2} - n^{2} = (n^{2} + 2n + 1) - n^{2} =

= n^{2}  + 2n + 1 - n^{2} = 2n + 1

For the sum of the integers we get:

n + (n + 1) = 2n + 1

Since 2n + 1 = 2n + 1, the difference of the squares of two consecutive positive integers is equal to the sum of the integers.

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Finding the common denominator and simplifying:

\frac{1}{\sqrt{x} - 1} - \frac{1}{\sqrt{x} + 1} = \frac{1}{\sqrt{x} - 1} \times \frac{\sqrt{x} + 1}{\sqrt{x} + 1} - \frac{1}{\sqrt{x} + 1} \times \frac{\sqrt{x} - 1}{\sqrt{x} - 1} =

= \frac{\sqrt{x} + 1}{(\sqrt{x} + 1)(\sqrt{x} - 1)} - \frac{\sqrt{x} - 1}{(\sqrt{x} + 1)(\sqrt{x} - 1)} = \frac{\sqrt{x} + 1}{x - 1} - \frac{\sqrt{x} - 1}{x - 1} =

= \frac{\sqrt{x} + 1 - (\sqrt{x} - 1)}{x - 1} = \frac{2}{x - 1}

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Let’s represent the integers as n, n + 1, and n + 2

(a)

The sum of the integers is:

n + (n + 1) + (n + 2) = 3n + 3

Factorising, we get that the sum is 3(n + 1), therefore the sum of three consecutive integers is divisible by 3.

(b)

The sum of the squares of the integers is:

n^{2} + (n + 1)^{2} + (n + 2)^{2} =

= n^{2} + (n^{2} + 2n + 1) + (n^{2} + 4n + 4)

= 3n^{2} + 6n + 5

Factorising 3 from the first two terms we get:

3(n^{2} + 2n) + 5

Since 3(n^{2} + 2n) is divisible by 3 while 5 is not divisible by 3, the sum of the squares of three consecutive integers is never divisible by 3.

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Let’s represent the integer as n.

Based on the information given in the question we can write:

\left ( n^{2} + \frac{1}{n} \right ) \times n  - 1= n^{3} + 1 - 1 = n^{3}

Since n is an even integer so divisible by 2, n^{3} is divisible by 8.

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Let’s represent the integers as n, n + 1, n + 2, and n + 3

The mean of the integers can be written as:

\frac{n +(n + 1)+(n + 2)+(n + 3)}{4} = \frac{4n + 6}{4} = \frac{2n + 3}{2}

The median of the four integers is:

\frac{(n + 1)+(n + 2)}{2} = \frac{2n + 3}{2}

Since \frac{2n + 3}{2} = \frac{2n + 3}{2}, the mean of four consecutive positive integers is equal to the median of these integers.

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