The following lemma may be used in this exercise. Lemma: If m is any integer, then m² +m is even. Proof: In case m is odd, then m² is odd and so m² + m is a sum of two odd integers, which is even. In case m is even, then m² is even and so m² + m is a sum of two even integers, which is even. By the parity principle, m is either even or odd, and in both cases m² + m is even. Consider the following statement. For every integer n 20, n(n² + 5) is divisible by 6. Prove the statement by filling in the blanks below. Proof (by mathematical induction): Let the property P(n) be the following sentence. The expression --Select-- ✓is divisible by. Show that P is true: P P() The expression --Select-- This is true because Select is the following sentence. Vis divisible by V-Select--V, which is divisible by Show that for each integer k z Let k be any integer with k2, and suppose that ---Select--- By definition of divisibility, this implies that there is an integer r such that k³ + Sk =()r, or, equivalently, K³= In the inductive step, we must show that P(k+ 1) is true. In other words, we must show that ---Select--- Now (k + 1){(k+ 1)² + 5) = (x + 1)((k² + ()*+1) + 5) = (x + 1)(k² + ( )k+6) if P(k) is true, then P(k+ 1) is true: Vis divisible by. (This is the inductive hypothesis.) =k³+ ()*² + ( )k+6 = (()r-5k) + 3k² + ()*+6 [by inductive hypothesis we know that there is an integer r such that k³=()r - Sk] )r + 3k² + ( )k+6 ])r +() (k² +k) +6 (Equation 1). By the lemma, k² + k is even, and so there is an integer s with k2 + k = 2s. Substituting into Equation 1 gives that (k+ 1)((k+ 1)² + 5) - ()r +() - 25+6 ()(r+s+1) By definition of divisibility and the fact that r + s + 1 is an integer, we can conclude that ---Select--- Therefore, P(k+ 1) is true. r- Sk. Vis divisible by . Vis divisible by