Soru

Zorluk: OrtaOdd and Even Integers (Parity)

If nn is any integer, which of the following expressions must be an even integer?

  1. 3n2+5n+23n^2 + 5n + 2Cevap
  2. B
    n2+n+1n^2 + n + 1
  3. C
    2n2+3n2n^2 + 3n
  4. D
    n3+2nn^3 + 2n
  5. E
    n2+3n^2 + 3

Cevap

The expression 3n2+5n+23n^2 + 5n + 2 must be an even integer for any integer nn.
The expression 3n2+5n+23n^2 + 5n + 2 can be factored as 3n(n+1)+23n(n+1) + 2. Since nn and n+1n+1 are consecutive integers, one of them must be even, making n(n+1)n(n+1) an even integer. Multiplying an even integer by 3 produces an even integer, and adding 2 maintains even parity. Therefore, this expression is guaranteed to be even for all integer values of nn.

Adım Adım Çözüm

1
Analyze the parity property of consecutive integers.
For any integer nn, one of the terms in the pair {n,n+1}\{n, n+1\} is even. Hence, the product n(n+1)n(n+1) is always an even integer.
The product of an even integer and any integer is always even.
2
Rewrite the target expression 3n2+5n+23n^2 + 5n + 2 to isolate the consecutive integer product.
3n2+5n+2=3n2+3n+2=3n(n+1)+23n^2 + 5n + 2 = 3n^2 + 3n + 2 = 3n(n+1) + 2.
Algebraic manipulation isolates known parity components.
3
Determine the parity of 3n(n+1)+23n(n+1) + 2.
Since n(n+1)n(n+1) is even, 3×even=even3 \times \text{even} = \text{even}. Then even+2=even\text{even} + 2 = \text{even}.
Multiplying an even integer by an odd integer yields an even integer, and adding an even integer preserves even parity.

Anahtar Kavram

Product of consecutive integers n(n+1)n(n+1) is always even; zero is an even integer; basic parity rules under addition and multiplication.
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