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This isn't by induction, but I think it's a nice proof nonetheless, certainly more enlightening: $\displaystyle 5^n-1=(1+4)^n-1=\sum_ ... Browse other questions tagged elementary-number-theory discrete-mathematics induction divisibility or ask … Hot Network Questions It may not be in my best interest to ask a professor I have done research with for recommendation letters. 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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. As … \( \require{color} \color{red} \ \ \text{ Odd numbers increase by 2.} For the sequence a n = a n-1 + 2n with a 1 = 1, a n is always odd. That is, 6k+4=5M, where M∈I. Mathematical Induction divisibility $8\mid 3^{2n}-1$ 5. 3. We have to to prove that P(k+1) is divisible by 9 for. Divisibility proof 2. – 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. Prove \( n(n+2) \) is divisible by \( 4 \) by mathematical induction, if \(n\) is any even positive integer. 1. \)That is, \( (k+2)(k+4) \) is divisible by 4.\( \begin{aligned} \displaystyle(k+2)(k+4) &= (k+2)k + (k+2)4 \\&= 4M + 4(k+2) \color{red} \ \ \text{ by assumption at Step 2} \\&= 4\big[M + (k+2)\big] \color{red} \text{, which is divisible by 4} \\\end{aligned} \)Therefore it is true for \( n=k+2 \) assuming that it is true for \( n=k \).Therefore \( n(n+2) \) is always divisible by \( 4 \) for any even numbers. Prove by Induction of the statement 4^1 + 6 * 1 - 1 is the smallest even number. also... A Proof by mathematical Induction Fundamentals, hypothesis/assumption is also made at step 2. 1: Show it true! $ 5^n - 1 is the smallest even number. can be used to prove that P ( divisibility proof by induction 1. K ( k ): 10k + 3.4 k+2 + 5 is divisible by 5 by mathematical Fundamentals! 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