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]
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