Harry | Maths Tutor
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Wednesday 17 December 2025 18:30:49 GMT
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☦︎光🇷🇺☦︎ :
n^2+5n = n(n+5). Either n is even, then n(n+5) is also even, or n is odd, then (n+5) is even, then n(n+5) is also even.
2025-12-18 07:46:48
192
nerisid :
Rewrite n^2 + 5n as n(n+1) + 4n and you’re done
2025-12-17 23:54:59
40
jonny21110 :
Proof should not be on the a level syllabus
2026-03-13 10:41:00
1
Peewee :
You forgot one case where n=1, since 2k and 2k+1 do not include 1 if k is in N.
2025-12-18 02:07:13
9
citharaaeternitaris :
I feel like n^2+5n =n*(n+5) should get you most of the way there. For n = 2k it's trivially even, for n = 2k+1 you have (2k+1)*(2k+1+5), so it's even because 2k+6 is even. Full algebraic expansion seems pointless.
2025-12-17 21:57:55
13
Amy :
Could you also do proof by induction ie n=1 is even, if we assume n=k is even then do n=k+1 and take the result for n=k you get 2k+6 which is even
2025-12-18 03:53:16
9
Fernando M :
i think you can also use partial sums here.
2025-12-18 08:18:11
0
declan :
Assume n is even. n^2 is always even. 5n is always even. Assume n is odd. n^2 is always odd. 5n always odd. odd + odd always = even.
2025-12-18 11:51:39
7
Gagas Math Nerd :
ok imma solve it and come back
2025-12-20 15:18:54
1
Christian :
Can I also prove that there exists no natural n such that n²+5n = 2n-1 by contradiction and thereby n must be even?
2025-12-18 12:28:29
1
Robbie Dexter :
I miss this part of learning math proving😍
2026-02-04 09:32:37
1
ellis :
wait hold it im on gcse and can do this what makes it so harc
2025-12-17 19:37:14
5
Fazlan Jay :
You are not checking for n=1
2026-01-22 14:05:26
0
Neka stani malo :
just prove for n = 1, then assume its even for n, and prove for n + 1 with that assumption, classic mathematical induction
2025-12-20 13:58:24
1
sam :
Easy, if n is odd then n = (2x+1) and n *n is odd so n²+5(2x+1) => (2a+1) + 10x+5 => 2a+10x+6 => where a, x and 6 are even so their sum is even because 2v + 2z = 2k
2025-12-20 13:02:49
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VAGABONDS56 :
this doesn't seem that bad it's just show the odd and even side and over conclusion this is similar to GCSE
2026-05-20 06:21:46
0
record of ragnarok fan :
its n(n+5) and then n is either even odd so n=2k or n=2k+1 if n=2k n(n+5) =2k(2k+5) which is even if n =2k+1 n(n+5)= (2k+1)(2k+6)=(2k+1)(2)(k+3) which is even
2025-12-20 15:41:02
0
Mathemagics :
n^2+5n=n^2+n (mod 2)=n(n+1) (mod 2) = 0 (mod 2) done
2025-12-20 20:13:47
0
GAV :
why can u also test for odd and even as case 3
2026-02-08 09:41:50
0
Dimee :
n=5
2025-12-20 08:51:39
0
່ :
isn’t this igcse level
2025-12-20 08:00:21
2
turgutmakbak :
I think the problem may be that people don’t know (or have prior experience) that this is a proof by cases. A way to avoid cases is a proof by contradiction. Suppose n²+5n is not even, then it’s odd. That means n(n+5) is odd, so both n and n+5 are odd. But if n+5 is odd, then n is even. So we get that n is both odd and even, which is impossible. So our initial assumption was incorrect, and thus n²+5n is even.
2025-12-19 18:07:08
0
Georg_xat :
you could prove that the equation is true for n=1, accept that it is even for n=k, prove that it's true for n=k+1 (k+1)^2+5(k+1)=k^2+2k+1+5k+5=(k^2+5k)+(2k+6) the first parenthesis is the equation for n=k (accepted as even) and the second parenthesis is also even 2(k+3) so the equation is giving always even number for any k of it is natural number
2025-12-18 08:20:18
0
drmolise :
Hey, in this one you didn't strictly complete the proof for all N. I give you 95%. Guess what you left out from you choices 2k and 2k+1 for k ê N?
2025-12-21 06:52:58
1
FUNBOY :
your explanation is good
2025-12-20 16:08:43
0
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