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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
YES
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20 with an exponent of 2 means 20 squared, or 20 times 20, or 400. An exponent of 3 means 20 cubed, or 20 times 20 times 20, or 8000. The exponents work as they do for any other base.
Your answer is not complete, as you need something after the last "plus". To give an answer that's still correct though, let's call it n: If: y = x2/(x2 + n) dy/dx = [2x(x2 + n) - (2x + dn/dx) ] / (x2 + n)2 = (2x3 + 2nx - 2x - dn/dx) / (x4 + 2nx2 + n2) = (2x3 + [2n - 2]x - dn/dx) / (x4 + 2nx2 + n2) If n is a constant, then you can remove the term dn/dx, and simplify the term [2n - 2]x