Prove by method of induction, for all $n \in N$: $(2^{3n}-1)$ is divisible by $7$.

Prove by method of induction, for all n ∈ N: (2³ⁿ-1) is divisible by 7.

EXERCISE 4.1 · Q10

n \in N

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Let $P(n):2^{3n}-1=7m$ for some $m\in N$ ; note $2^{3n}=8^n$ . Base: $n=1$ : $8-1=7=7(1)$ ; holds. Hypothesis: assume $8^k-1=7a$ , i.e. $8^k=7a+1$ . Step: $8^{k+1}-1=8\cdot8^k-1=8(7a+1)-1=56a+8-1=56a+7=7(8a+1)$ , a multiple of 7. **Concl …

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Q.Prove by method of induction, for all $n \in N$ : $(2^{3n}-1)$ is divisible by $7$ .

Concept understanding — Principle of Mathematical Induction

Q.Prove by method of induction, for all n∈Nn \in Nn∈N: (23n−1)(2^{3n}-1)(23n−1) is divisible by 777.

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Q.Prove by method of induction, for all $n \in N$ : $(2^{3n}-1)$ is divisible by $7$ .

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