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16 Φεβ 2016 · Prove the following statement by mathematical induction: n(n2 + 5) is divisible by 6 for all integer n ≥ 1. My attempt: Let the given statement be p (n). (1) 1(12 + 5) =6 Hence, p (1) is true. (2) Suppose for all integer k ≥ 1, p (k) is true. That is, k(k2 + 5) is divisible by 6. We must show that p (k+1) is true.
18 Ιαν 2020 · Learn how to use mathematical induction to prove a divisibility statement in this video tutorial.
3 Σεπ 2020 · 1 Answer. +1 vote. answered Sep 3, 2020 by Shyam01 (49.0k points) selected Sep 4, 2020 by Chandan01. Best answer. According to the question, P (n) = n (n2 + 5) is divisible by 6. So, substituting different values for n, we get, P (0) = 0 (02 + 5) = 0 Which is divisible by 6. P (1) = 1 (12 + 5) = 6 Which is divisible by 6.
18 Ιαν 2024 · In the Details mode, you can understand why a number between 2 and 13 is (or is not) a divisor of a given integer. The calculator will display the appropriate divisibility rule and will show you how to apply it. A number is divisible by 10 if and only if its last digit is 0.
28 Αυγ 2019 · Best answer. Let (P): n(nn + 5) ( P): n ( n n + 5) is divisible by 6, for each natural number n. Step I We observe that P (1) is true. P (1): 1(12 + 5) = 6 P ( 1): 1 ( 1 2 + 5) = 6, which is divisible by 6. Step II Now, assume that P (n) is true for n=k. P (k): k(k2 + 5) P ( k): k ( k 2 + 5) is divisible by 6. ∴ k(k2 + 5) = 6q ∴ k ( k 2 + 5) = 6 q.
4 Νοε 2012 · Homework Helper. 3,802. 95. No no no... You shouldn't instantly jump to f (k+1)-f (k) to try and prove this.
Divisibility Rules: 2, 3, 4, 5, 6, 9, and 10. A number [latex]a[/latex] is divisible by the number [latex]b[/latex] if [latex]a \div b[/latex] has a remainder of zero ([latex]0[/latex]). For example, 15 divided by 3 is exactly 5 which implies that its remainder is zero. We then say that 15 is divisible by 3.
