Question

Proof: If n is odd, then 4n^2 +7n +6 is odd.

Answer 1

take a random odd number

4(3)² + 7(3) + 6
4(9) + 21 + 6
36 + 21 + 6
36 + 27
63

so yes

Answer 2

could have tried it with 1 too

4(1)² + 7(1) + 6
4(1) + 7 + 6
4 + 7 + 6
11 + 6
17

so again yes

Answer 3

I need it to be proven for all n, not just any arbitrary odd #

Answer 4

Both 3 and 1 are not random :)

A random odd number has the for m 2k+1, where k is an integer.

Answer 5

exactly ZeHanz, I just can't figure out where to go from there

Answer 6

So put 2k+1 into the thing, and see if it holds:

Expand: 4(2k+1)² +7(2k+1) +6

Answer 7

This gives, after some calculating: 16k²+30k+17
Now this is the sum of three terms:
16k²: even.
30k: even.
17: uneven.
So the sum is uneven.

Answer 8

uneven=odd in my world ;D

Answer 9

Thank you, This is exactly what I needed.

Answer 10

Another way is to also use the properties for example. Even times Even equals to even and so on. Sorry I didn't memorize them but hoped I helped.

Answer 11

YW!