Given that m is an integer,
(a) write down expressions for the next two odd numbers after 2n - 1,
b) © find, in its simplest form, the expression for the sum of these three odd numbers,
(©) explain why the sum is a multiple of 3,
(c) find, in its simplest form, an expression for the sum of the squares of these three odd
numbers.
a) they would have to be 2 units apart, so:
2n-1, 2n+1, 2n+3
b) sum of those = 2n-1 + 2n+1 + 2n+3 = 6n +3
we can factor out 3 to get 3(2n+1) which is a multiple of 3, since 3 is a factor
c) sum of squares = (2n-1)^2 + (2n+1)^2 + (2n+3)^2
= 4n^2 - 4n + 1 + 4n^2 + 4n + 1 + 4n^2 + 12n + 9
= 12n^2 + 12n + 11
(a) To find the next two odd numbers after 2n - 1, we need to find the odd numbers that come after the given expression.
The next odd number after 2n - 1 is obtained by adding 2 to 2n - 1. So, the first odd number is 2 + (2n - 1) = 2n + 1.
To find the next odd number after that, we add an additional 2 to 2n + 1. So, the second odd number is 2 + (2n + 1) = 2n + 3.
Therefore, the next two odd numbers after 2n - 1 are 2n + 1 and 2n + 3.
(b) To find the sum of the three odd numbers, we add them together. The three odd numbers are 2n - 1, 2n + 1, and 2n + 3.
Sum = (2n - 1) + (2n + 1) + (2n + 3)
= 2n - 1 + 2n + 1 + 2n + 3
= 6n + 3
Simplifying this expression gives us the sum of the three odd numbers: 6n + 3.
(©) The sum 6n + 3 is a multiple of 3 because it can be divided evenly by 3 without leaving any remainder.
To explain why the sum is a multiple of 3, we can consider the fact that any multiple of 3 can be written as 3k, where k is an integer. In our case, the sum 6n + 3 can be written as 3(2n + 1), where (2n + 1) is an integer.
Since 3 can divide evenly into 3(2n + 1) without leaving a remainder, the sum 6n + 3 is indeed a multiple of 3.
(c) To find the sum of the squares of the three odd numbers, we need to square each number, and then add them together.
The three odd numbers are 2n - 1, 2n + 1, and 2n + 3.
Sum of squares = (2n - 1)^2 + (2n + 1)^2 + (2n + 3)^2
Expanding and simplifying each term:
(2n - 1)^2 = (2n - 1)(2n - 1) = 4n^2 - 4n + 1
(2n + 1)^2 = (2n + 1)(2n + 1) = 4n^2 + 4n + 1
(2n + 3)^2 = (2n + 3)(2n + 3) = 4n^2 + 12n + 9
Sum of squares = (4n^2 - 4n + 1) + (4n^2 + 4n + 1) + (4n^2 + 12n + 9)
= 12n^2 + 12n + 11
Therefore, the sum of the squares of the three odd numbers is 12n^2 + 12n + 11.