Step 3: Show it is true for n=k+1. \( \require{color} \color{red} \ \ \text{ 1 is the smallest odd number.} Answer Save. Divisibility proofs by induction. \( \require{color} \color{red} \ \ \text{ Odd numbers increase by 2.} In this post, we will explore mathematical induction by understanding the nature of inductive proof, including the ‘initial statement’ and the inductive step. Prove \( 4^n + 5^n + 6^n \) is divisible by \( 15 \) by mathematical induction, where \(n\) is odd integer. Step 1: Show it is true for n=0. In this post, we will explore mathematical induction by understanding the nature of inductive proof, including the ‘initial statement’ and the inductive step. I've recently come across a divisibility problem that I am unable to solve. 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Mathematical Induction Divisibility Problem. Prove \( 5^n + 2 \times 11^n \) is divisible by \( 3 \) by mathematical induction. Mathematical Induction question: Prove divisibility by $4$ of $5^n + 9^n + 2$ 4. mathematical induction for divisibility: Is this one a valid proof? For each of the statements below, the corresponding activity will lead you to explore a proof by induction of the statement. Induction proof, divisibility. 6k+1+4=6×6k+4=6(5M–4)+46k=5M–4by Step 2=30M–20=5(6M−4),which is divisible by 5 Therefore it is true for n=k+1 assuming that it is true for n=k. \)That is, \( 4^{k+2} + 5^{k+2} + 6^{k+2} \) is divisible by \( 15 \).\( \begin{aligned} \displaystyle \require{color}4^{k+2} + 5^{k+2} + 6^{k+2} &= 4^k \times 4^2 + 5^k \times 5^2 + 6^k \times 6^2 \\&= (15M – 5^k – 6^k) \times 4^2 + 5^k \times 5^2 + 6^k \times 6^2 \\&= 240M – 16 \times 5^k – 16 \times 6^k + 25 \times 5^k + 36 \times 6^k \\&= 240M + 9 \times 5^k + 20 \times 6^k \\&= 240M + 9 \times 5 \times 5^{k-1} + 20 \times 6 \times 6^{k-1} \\&= 240M + 45 \times 5^{k-1} + 120 \times 6^{k-1} \\&= 15\big[16M + 3 \times 5^{k-1} + 8 \times 6^{k-1}\big], \text{ which is divisible by 15} \\\end{aligned} \)Therefore it is true for \( n=k+2 \) assuming that it is true for \( n=k \).Therefore \( 4^n + 5^n + 6^n \) is always divisible by \( 15 \) for all odd integers. 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