Answer: [tex](a)\ n+1[/tex] Step-by-step explanation: Given [tex]n \to even[/tex] Required Which cannot be even To do this, we assume an even number for n; Assume that [tex]n = 4[/tex] So, we have: [tex](a)\ n+1[/tex] [tex]n +1 = 4 + 1 = 5[/tex] — this is odd [tex](b)\ n+2[/tex] [tex]n +2 = 4+2 = 6[/tex] —- this is even [tex](c)\ 2n[/tex] [tex]2n = 2 * 4 = 8[/tex] — this is even [tex](d)\ n^2[/tex] [tex]n^2 = 4^2 = 16[/tex] — this is even Hence, (a) can’t be even Log in to Reply

Answer:[tex](a)\ n+1[/tex]

Step-by-step explanation:Given[tex]n \to even[/tex]

RequiredWhich cannot be even

To do this, we assume an even number for n;Assume that [tex]n = 4[/tex]

So, we have:[tex](a)\ n+1[/tex]

[tex]n +1 = 4 + 1 = 5[/tex] — this is odd

[tex](b)\ n+2[/tex]

[tex]n +2 = 4+2 = 6[/tex] —- this is even

[tex](c)\ 2n[/tex]

[tex]2n = 2 * 4 = 8[/tex] — this is even

[tex](d)\ n^2[/tex]

[tex]n^2 = 4^2 = 16[/tex] — this is even

Hence, (a) can’t be even