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Zorluk: KolayOdd and Even Integers (Parity)

For any integer nn, the expression n2+n+7n^2 + n + 7 is an odd integer.

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Cevap

The statement is true because n(n+1)n(n + 1) is the product of two consecutive integers, which is always even. Adding the odd constant 77 to an even integer yields an odd integer for all integer values of nn.
The statement is true because n2+n=n(n+1)n^2 + n = n(n + 1) represents the product of two consecutive integers and is therefore always even. Adding 77 (an odd integer) to an even integer always yields an odd integer regardless of whether nn is positive, negative, or zero.

Adım Adım Çözüm

1
Factor the algebraic terms n2+nn^2 + n.
n2+n+7=n(n+1)+7n^2 + n + 7 = n(n + 1) + 7
Factoring out nn reveals the product of two consecutive integers.
2
Determine the parity of the product of two consecutive integers n(n+1)n(n + 1).
n(n+1)n(n + 1) is always even.
Of any two consecutive integers, exactly one is even. The product of an even integer and any integer is always even.
3
Determine the overall parity of the expression by adding 77.
even+7=odd\text{even} + 7 = \text{odd}
Adding an odd integer to an even integer results in an odd integer.

Anahtar Kavram

The product of two consecutive integers is always even. Adding an odd integer to an even integer results in an odd integer.
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