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Let ${a_n}$ be the sequence given by $a_1 = 3$ and $a_{n+1} = 2a_n + 5$. Use induction to prove that $a_n > 2^n$ for all $n \in \mathbb{N}$

Attempt:

For $n = 1,$ we have $3>2$ so the base case holds.

Assume $a_k > 2^k$ for some k

For $n = k+1$ we have:

$a_{k+1} = 2 a_{k} + 5$

John
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2 Answers2

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If $a_k > 2^k$, then $$2a_k + 5 > 2 a_k > 2 \cdot 2^k$$ as desired.

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You could infact obtain $a_n$ with reasonable ease; Note that $$a_{n+1} + 5 = 2(a_n+5)$$ Hence, we have $$a_{n+1} + 5 = 2(a_n+5) = 2(2(a_{n-1}+5)) = 2^k (a_{n-k+1}+5) = 2^n(a_1 + 5)$$ Hence, we get $$a_{n+1} +5 = 2^n(3+5) = 2^{n+3} \implies a_n = 2^{n+2} - 5$$ Now show that for $n \geq 1$, we have $$2^{n+2} - 5 > 2^n$$