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Chapter 1: Problem 8

Prove that \(5^{n}-4 n-1\) is divisible by 16 for all \(n \in \mathbb{N}\).

Short Answer

Expert verified
By the inductive method, every value of \(5^{n} - 4n - 1\) where \(n\) is a natural number can be simplified to a multiple of 16, so it is divisible by 16.

Step by step solution

01

Base Case

First, validate the base case. Substitute \(n = 1\) into the formula, which gives \(5^1 - 4*1 - 1 = 0\). Since this result is divisible by 16, the base case holds true.
02

Inductive Step - Assumption

Next, make the inductive hypothesis. Assume that the formula holds true for some arbitrary \(n = k\). In this case, the statement is \(5^k - 4k - 1\) is divisible by 16.
03

Inductive Step - Validation

Then, demonstrate that the statement should thus also hold true for \(n = k + 1\). Confirm this by showing that \(5^{k+1} - 4(k+1) - 1\) can be reduced to a multiple of 16, given our assumption.
04

Reduction and Simplification

To continue the last step, first expand and simplify \(5^{k+1} - 4(k+1) - 1\): this equals \(5*5^k - 4k - 4 - 1\), or \(5*5^k - 4k - 5\). Then, restyle this as \(5(5^k - 1) - 4(k + 1)\). We know from our assumption that \(5^{k} - 1\) is a multiple of 16, so \(5(5^k - 1)\) must be as well. As for \(-4(k + 1)\), this is also a multiple of 16, because -4 times any integer equals a multiple of 16. Thus, the entire expression \(5*5^{k} - 4(k + 1)\) is a multiple of 16.

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