Prove by mathematical induction that (2C2)+(3C2)+(4C2)+...+(n-1C2) = (nC3).
Firstly, show the equation is true for n = 3 (as this is the samllest n that nC3 is defined): LHS = (2C2) = 1 = (3C3) = RHStherefore, true for n=3.
Then assume true for n = k:(2C2)+(3C2)+(4C2)+...+(k-1C2) = (kC3).
Concider n = k-1:(2C2)+(3C2)+(4C2)+...+(k-1C2)+(kC2) = (kC3)+(kC2) = [k!/(k-3)!3!] + [k!/(k-2)!2!] = (k!/3!)[(1/(k-3)!)+3/(k-2)!] = (k!/3!)[(k-2+3)/(k-2)!] = (k!/3!)[(k+1)/(k-2)!] = [(k+1)!/3!(k-2)!] = (k+1)C3
Equation is true for n = 3. If true for n = k, it is true for n = k+1. Therefore the equation is true for all n >= 3 by induction.
HX
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