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.
CloseTo 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:
\[ (3n-2)^3 + (3n+2)^3 = 54n^3 + 72n \]
Closea)
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: \[ (-2)^3 + 9 \cdot (-2)^2 + 27 \cdot (-2) + 27 = -8 + 36 - 54 + 27 = 1 \]
CloseWe 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.
Closea)
\[ \frac{16x^2 - 4}{4x^2 + 6x + 2} = \frac{4x - 2}{x + 1} \]
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: \( (x+1) \neq 0 \)
\[ (4x+2)(x+1) \neq 0 \]
\[ (x+1) \neq 0 \]
\[ x \neq -1, \ x \neq -\frac{1}{2} \]
CloseLet'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.
CloseSince 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 \( b \).
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.
Close