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. Thanks for any help. Prove that 4^n + 6n -1 is diisible by 9 for all n>=1. Step 1: Show it is true for n=0. 10 years ago. I know that most of these types of problems have fairly straightforward proof-by-induction solutions -- but for this particular problem, I don't know how to finish the inductive step. Let us assume that P(n) is true for some natural number n = k. or K3 â 7k + 3 = 3m, mâ N         (i). Randy P. Lv 7. â 7n + 3 is divisible by 3, for all natural numbers n. if you need any other stuff in math, please use our google custom search here. Step 1:  Show it is true for $$n=1$$. For the sequence a n = a n-1 + 2n with a 1 = 1, a n is always odd. If so why? I've recently come across a divisibility problem that I am unable to solve. Prove 6n+4 is divisible by 5 by mathematical induction. Step 3: Show it is true for n=k+1. 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Divisibility proof 2. 60+4=5, which is divisible by 5 Step 2: Assume that it is true for n=k. Now, we have to prove that P(k + 1) is true. I need to prove that 7^n + 4^n +1 is divisible by 6 using induction, I habe gotten as far as the last step of n=k+1 which I am stuck on. Mathematical Induction: Divisibility This is part of the HSC Mathematics Extension 1 course under the topic Proof by Mathematical Induction. Step 1:  Show it is true for $$n=0$$.$$6^0 + 4 = 5$$, which is divisible by $$5$$Step 2:  Assume that it is true for $$n=k$$.That is, $$6^k + 4 = 5M$$, where $$M \in I$$.Step 3:  Show it is true for $$n=k+1$$.That is, $$6^{k+1} + 4 = 5P$$, where $$P \in I$$.\begin{aligned} \displaystyle \require{color}6^{k+1} + 4 &= 6 \times 6^k +4 \\&= 6 (5M – 4) + 4 \ \ \ \color{red} 6^k = 5M – 4 \ \ \ \ \text{ by Step 2} \\&= 30M – 20 \\&= 5(6M-4), \text{ which is divisible by 5} \\\end{aligned}Therefore it is true for $$n=k+1$$ assuming that it is true for $$n=k$$.Therefore $$6^n + 4$$ is always divisible by $$5$$. The base case shows that the statement is true for the first natural number, and the induction step shows that the statement is true for the next one. I've recently come across a divisibility problem that I am unable to solve. Hence, P(l) is true. 2. 2. Use induction to prove that n3 â 7n + 3, is divisible by 3, for all natural numbers n. 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