Prove using induction that $2^{3n-1} + 3$ is divisible by 7 for all $n \\in \\mathbb{N}$. - Leaving Cert Mathematics - Question 4(a) - 2021
Question 4(a)
Prove using induction that $2^{3n-1} + 3$ is divisible by 7 for all $n \\in \\mathbb{N}$.
Worked Solution & Example Answer:Prove using induction that $2^{3n-1} + 3$ is divisible by 7 for all $n \\in \\mathbb{N}$. - Leaving Cert Mathematics - Question 4(a) - 2021
Step 1
Base Case: $n = 1$
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Answer
To establish the base case, we evaluate the expression when n=1:
2^{3(1)-1} + 3 &= 2^2 + 3 \\
&= 4 + 3 \\
&= 7.
\end{align*}$$
We can see that 7 is divisible by 7, thus the base case holds.
Step 2
Inductive Hypothesis: Assume for $n = k$
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Answer
Assume that the statement holds for some integer k. That is, we assume:
P(k):23k−1+3extisdivisibleby7.
This means there exists an integer m such that:
23k−1+3=7m.
Step 3
Inductive Step: Show for $n = k + 1$
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Answer
We need to show that:
$$P(k + 1): 2^{3(k+1)-1} + 3 ext{ is divisible by } 7.$$$$
We compute P(k+1):
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Answer
Since both the base case and the inductive step have been established, we conclude that by mathematical induction, the statement is true for all ninmathbbN.
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