Question
Write an expression for the sum of 1 and twice a number n. If you let n be any odd number, will the result always be an odd number?
Answers
GPT-4o mini
To create the expression for the sum of 1 and twice a number \( n \), we can write it as:
\[
1 + 2n
\]
Now, let's analyze whether this expression is always odd when \( n \) is an odd number.
An odd number can be represented as \( n = 2k + 1 \), where \( k \) is an integer. If we substitute this representation into our expression:
\[
1 + 2n = 1 + 2(2k + 1) = 1 + 4k + 2 = 4k + 3
\]
The term \( 4k \) is clearly even, and adding 3 (which is odd) to an even number results in another odd number. Therefore, \( 4k + 3 \) is guaranteed to be odd.
Thus, when \( n \) is any odd number, the expression \( 1 + 2n \) will always yield an odd result.
\[
1 + 2n
\]
Now, let's analyze whether this expression is always odd when \( n \) is an odd number.
An odd number can be represented as \( n = 2k + 1 \), where \( k \) is an integer. If we substitute this representation into our expression:
\[
1 + 2n = 1 + 2(2k + 1) = 1 + 4k + 2 = 4k + 3
\]
The term \( 4k \) is clearly even, and adding 3 (which is odd) to an even number results in another odd number. Therefore, \( 4k + 3 \) is guaranteed to be odd.
Thus, when \( n \) is any odd number, the expression \( 1 + 2n \) will always yield an odd result.