Proof by induction

[Oaky]

(this may be in the wrong thread, apologies!)

Prove by induction that the given formula is true for every positive integer n.

P(n) says that
2+4+6+ ... +2n = n(n+1)


- Firstly I proved that P(1) is true
- Next I assumed that P(k) is true, ie. 2+4+6+ ... +2k = k(k+1)
- So I need to show, using this assumption, that P(k+1) is true

LHS P(k+1)
= 2+4+ ... + 2(k+1)
= 2+4+ ... + 2k + 2
= [2+4+ ... +2k] + 2
from P(k)
= k(k+1) +2
= k²+ k + 2

RHS P(k+1)
= (k+1)(k+2)
= k²+ 3k + 2

Now obviously LHS P(k+1) doesn't equal RHS P(k+1).
So where am I going wrong? I have a feeling it comes with the grouping in the 3rd step of LHS P(k+1) but don't know what's invalid.

Thanks so much guys

 

[pka]

(this may be in the wrong thread, apologies!)

Prove by induction that the given formula is true for every positive integer n.

P(n) says that
2+4+6+ ... +2n = n(n+1)

LHS P(k+1)
= 2+4+ ... +2k+ 2(k+1)
= [2+4+ ... +2k] + 2k+2
from P(k)
= k(k+1) +2k+2
= k²+ k + 2k+2

See my corrections above.