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I think that you're doing the same thing that I did for IB. Here's what I think you're asking for :

0, 2, 6, 12, 20, 30, 42...

c₁ = 0

c₂ = 2

c₃ = 6 (c₂ + 4 = 2 + 4 = 6)

c₄ = 12 (c₃ + 6 = 2 + 4 + 6 = 12)

c₅ = 20 (c₄ + 8 = 2 + 4 + 6 + 8 = 20)

c₆ = 2 + 4 + 6 + 8 + 10

etc...

dn = (n/2) <2c₁ + (n-1) 2>

dn = (n/2) <2 (2) + (n-1) 2>

dn = (n/2) (4 + 2n - 2)

dn = (n/2) (2 + 2n)

dn = (2n/2) + (2n²/2)

dn = n + n²

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15y ago

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