Answered: Discrete Mathematics: 3. Prove: 7n -1…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Discrete Mathematics:

3. Prove: 7_{n -1} is divisible by 3 for n ³ 0.

Expert Solution

Step 1

We need to prove that the number $7^{n} - 1$ is divisible by 3 for $n > 0$ . We do this as follows:

Case-1: n=1

We have $7^{n} - 1 = 7 - 1 = 6 = 2 \times 3$ which implies that for n=1, the given number is divisible by 3.

Case-2: $n > 1$

We can write the given number as ${(6 + 1)}^{n} - 1$ . Now, we use the binomial expansion i.e., ${(a + b)}^{n} = {}^{n}C_{0} a^{n} + {}^{n}C_{1} a^{1} b^{n - 1} + {}^{n}C_{2} a^{2} b^{n - 2} + . . . + {}^{n}C_{n} b^{n}$ for the term ${(6 + 1)}^{n}$ with a=6 and b=1.

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