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EXERCISES - MATHEMATICAL INDUCTION

nΣk=0 qk = (1-qn+1)/(1-q), ∀ n ∈ ℕ, q ∈ ℝ, q ≠ 1; q = 2, 1+2+4+8+16+32; Basis of induction: n = 0, 0Σk=0 qk = (1-q)/(1-q), q0 = 1, 1 = 1 true; Inductive step - Hypothesis: nΣk=0 qk = (1-qn+1)/(1-q); Inductive step - Thesis: n+1Σk=0 qk = (1-q(n+1)+1)/(1-q) = (1-q(n+2))/(1-q); n+1Σk=0 qk = nΣk=0 qk + n+1Σk=n+1 qk = (1-qn+1)/(1-q) + qn+1 = (1-qn+1+(1-q)qn+1)/(1-q) = (1-qn+1+qn+1-qn+2)/(1-q) = (1-qn+2)/(1-q)

The sum of the internal angles of a convex polygon of n+2 sides is equal to n flat angles (n·π); Basis of induction: n = 1, a convex polygon of n+2 sides with n=1 is a triangle, the sum of the interior angles of a triangle is π, 1·π = π, true; Inductive step - Hypothesis: the sum of the internal angles of a convex polygon of n+2 sides is n·π; Inductive step - Thesis: the sum of the internal angles of a convex polygon of (n+1)+2 = n+3 sides is (n+1)π; Considering a convex polygon of 5 sides, n = 2, n+3 sides convex polygon, it can be divided in a triangle and a quadrangle; the triangle is n+1 sides and the sum of the angles is π; the quadrangle is n+2 sides and the sum of the angles is n·π; angles of the triangle (π) + angles of the quadrangle (n·π) = π + n·π = (n+1)π