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For any integer \newlinenn, which of the following are always\newlinea) two consecutive even numbers?\newlineb) two consecutive odd numbers?\newline\newlinenn and \newlinen+2n+2\newlinen1n-1 and \newlinen+1n+1\newline2n2n and \newline2n+22n+2\newline\newline2n2n and \newline2n+12n+1\newline2n12n-1 and \newline2n+12n+1\newline2n+12n+1 and \newline2n+22n+2

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Q. For any integer \newlinenn, which of the following are always\newlinea) two consecutive even numbers?\newlineb) two consecutive odd numbers?\newline\newlinenn and \newlinen+2n+2\newlinen1n-1 and \newlinen+1n+1\newline2n2n and \newline2n+22n+2\newline\newline2n2n and \newline2n+12n+1\newline2n12n-1 and \newline2n+12n+1\newline2n+12n+1 and \newline2n+22n+2
  1. First Option Analysis: Let's consider the first option: nn and n+2n+2. If nn is an even number, then n+2n+2 is also even because adding 22 to an even number results in another even number. Similarly, if nn is an odd number, then n+2n+2 is also odd because adding 22 to an odd number results in another odd number. Therefore, nn and n+2n+2 are either both even or both odd, depending on the parity of nn.
  2. Second Option Evaluation: Now let's consider the second option: n1n-1 and n+1n+1. If nn is an even number, then n1n-1 is odd and n+1n+1 is odd because subtracting 11 from an even number results in an odd number and adding 11 to an even number also results in an odd number. If nn is an odd number, then n1n-1 is even and n+1n+1 is even because subtracting 11 from an odd number results in an even number and adding 11 to an odd number also results in an even number. Therefore, n1n-1 and n+1n+1 are always consecutive odd or even numbers, but they are not the same parity as each other.
  3. Third Option Assessment: Next, let's evaluate the third option: 2n2n and 2n+22n+2. Since 2n2n is always even for any integer nn, and adding 22 to an even number results in another even number, 2n+22n+2 is also even. Therefore, 2n2n and 2n+22n+2 are two consecutive even numbers.
  4. Fourth Option Examination: Now let's look at the fourth option: 2n2n and 2n+12n+1. Since 2n2n is always even for any integer nn, and adding 11 to an even number results in an odd number, 2n+12n+1 is odd. Therefore, 2n2n and 2n+12n+1 are not two consecutive even or odd numbers because they have different parities.
  5. Fifth Option Consideration: Let's consider the fifth option: 2n12n-1 and 2n+12n+1. Since 2n2n is always even for any integer nn, subtracting 11 from an even number results in an odd number, so 2n12n-1 is odd. Similarly, adding 11 to an even number results in an odd number, so 2n+12n+1 is also odd. Therefore, 2n12n-1 and 2n+12n+1 are two consecutive odd numbers.
  6. Last Option Review: Finally, let's evaluate the last option: 2n+12n+1 and 2n+22n+2. Since 2n+12n+1 is always odd for any integer nn, and adding 11 to an odd number results in an even number, 2n+22n+2 is even. Therefore, 2n+12n+1 and 2n+22n+2 are not two consecutive even or odd numbers because they have different parities.

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