Question

Which of the following is odd for every integer n ?
a. 2011n
b. n2 + 2011
c. n
d. 2n2 + 2011

Answer 1

Answer:

$n = odd \: number = 3$

$2 {n}^{2} + 2011 = 2027$

$n = even \: number = 2$

$2 {n}^{2} + 2011 = 2019$

SO THE VALUE OF 2n²+2011 IS ALWAYS ODD

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Answer 2

Answer:

n2+2011 is the answer

Step-by-step explanation: