I am trying to prove this statement, but I'm not sure where to go from here. Is don't think this is sufficiently reduced to conclude the statement is true, but I'm not positive.
$k ≥ 2$
Can I further simplify this?
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$\begingroup$ Is $k\in \mathbb{Z^+}$? $\endgroup$– foshoFeb 5, 2016 at 18:19
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$\begingroup$ Yes, $k$ is an integer $\endgroup$– 123Feb 5, 2016 at 18:21
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$\begingroup$ Do you mean $k = 1,2,3,...?$ $\endgroup$– foshoFeb 5, 2016 at 18:24
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$\begingroup$ Since $5^k \cdot 5 = 5^{k+1}$ and $6^k \cdot 6 = 6^{k+1}$, proving the result for all $k > 0$ is equivalent to prove that $5^k + 9 < 6^k$ for all $k > 1$. $\endgroup$– J.-E. PinFeb 5, 2016 at 18:26
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3 Answers
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Hint:
It's not hard to check that $6^{k+1}-{5}^{k+1}$ is increasing and $(6\cdot6-5\cdot 5) \geq 9$
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$$ 6^{k+1}-{5}^{k+1} = (6-5)(6^k + \cdots + 5^k) = 6^k + \cdots + 5^k \ge 6^k + 5^k \ge 6+5 =11 > 9 $$ for $k\ge 1$.
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You can prove it by induction suppose $5^{k+1} + 9 < 6^{k+1}$ that this is correct and you should prove it for $k+1$ so using the induction assumption the left side is smaller so what it would be if you multiply the part of left side by another $5$ and the right side by another $6$? the left side will be still smaller?
