∑i=1nfi2=fnfn+1sum from i equals 1 to n of f sub i squared equals f sub n f sub n plus 1 end-sub Step-by-Step Induction Proof .The base case holds. Inductive Step: Assume the formula holds for . We must show it holds for
of real numbers is defined as a if, for all indices , the following inequality holds: stefani_problem_stefani_problem
∑i=1k+1fi2=(∑i=1kfi2)+fk+12sum from i equals 1 to k plus 1 of f sub i squared equals open paren sum from i equals 1 to k of f sub i squared close paren plus f sub k plus 1 end-sub squared Substitute the inductive hypothesis: ∑i=1nfi2=fnfn+1sum from i equals 1 to n of
Assuming the property is false and showing this leads to an impossibility. Contraposition: Proving "If not B, then not A." Contraposition: Proving "If not B, then not A
Look into Monge Arrays to see how these "Gnome" properties allow for faster shortest-path algorithms in geometric graphs.