Induction can generally just go die in a hole now

2 + 4 + 6 + ... + 2n = n(n + 1)
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Does it work for 1?
2 = 1(2) Yes!
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Assuming 2 + 4 + 6 + ... + 2n = n(n + 1) is true:
2 + 4 + 6 + ... + 2(n + 1) = (n + 1)((n + 1) + 1)
2 + 4 + 6 + ... + 2n + 2 = (n + 1)(n + 2)
    ==================== <- this bit equals n(n + 1) by assumption
2 + n(n + 1) = (n + 1)(n + 2)
n² + n + 2 =/= n ² + 3n + 2
NO YOU IDIOT THAT'S NOT WHAT'S SUPPOSED TO HAPPEN DAMNIT