Mathematical induction -Split -

[lc312]

Hi, i have problem with solving this issue, and the question is: "Prove that for every odd ? ∈ ℕ number ?
2 + 2? + 3 is even". I mean I did solve that but I am not shure if that is correct.
Here is the picture how i solved it.
Thank you.

 

[JeffM]

Thank you very much for showing your work. Sad to say, only part of it is visible to me. Therefore I at least cannot comment until you either post a complete and legible picture or (better) write out what you did in a post.

 

[lc312]

n^2+2n+3

Part 1.
n=1
1^2+2*1+3
1+2+3=6 =>3*2=6

Part 2.
n=k
k^2+2k+3=2*a, a ∈ ℕ

Part 3
n=k+1
(k+1)^2+2*(k+1)+3
k^2+2k+1+2k+2+3
(k^2+2k+3)+1+2k+2
2a+2*(k+1)
2*(a+k+1)


I dont know if this is the right solution but i would be verry grateful if you can help solv this issue.
I dont have any ideas anymore.
Many thanks.

 

[HallsofIvy]

If I were your teacher I would ask for more "explanation"

n^2+2n+3

Part 1.
n=1
1^2+2*1+3
1+2+3=6 =>3*2=6

Therefore the statement is true for n= 1

Part 2.

Assume that for

n=k
k^2+2k+3=2*a, a ∈ ℕ

Part 3

Then, for

n=k+1
(k+1)^2+2*(k+1)+3
k^2+2k+1+2k+2+3
(k^2+2k+3)+1+2k+2

What happened to "+3"?

2a+2*(k+1)

Neither "1+ 2k+ 2" nor "1+ 2k+ 2+ 3" is equal to 2(k+1).
1+ 2k+ 2= 2k+ 3 is odd and 1+ 2k+ 2+ 3= 2k+ 6= 2(k+ 3).

2*(a+k+1)

No, it is 2(a+k+3)

I dont know if this is the right solution but i would be verry grateful if you can help solv this issue.
I dont have any ideas anymore.
Many thanks.

 

[Deleted member 4993]

n^2+2n+3

Part 1.
n=1
1^2+2*1+3
1+2+3=6 =>3*2=6

Part 2.
n=k
k^2+2k+3=2*a, a ∈ ℕ

Part 3
n=k+1
(k+1)^2+2*(k+1)+3
k^2+2k+1+2k+2+3
(k^2+2k+3)+1+2k+2
2a+2*(k+1)
2*(a+k+1)


I dont know if this is the right solution but i would be verry grateful if you can help solv this issue.
I dont have any ideas anymore.
Many thanks.

If "k" is an odd number - the next 'odd' number is "k +2"

so you need to check now:

(k+2)^2 + 2*(k+2) + 3

continue.....