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So from what I've found using excel this inequality wholes true where n > 2 where n is in naturals

I have 2 questions. How would I do this not using excel? Would the easiest way be to use a calculator? I dont understand how else I would solve this equation (by maybe setting them equal and coming up with an answer between 2 and 3 and then rounding up).

In adition, I'm pretty sure the way to solve this is by using induction. So for my base case let N = 3. Then I will let n = k + 1 and need to show that 5^(k+1) + 6^(k+1) > 7^(k+1).

I can manipulate this into saying (5^k * 5) + (6^k * 6) < (7^k * 7). But where would I go from here? I know I need to somehow use my IH but don't quote know how.

Thanks for the help.

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  • $\begingroup$ probably the best way to come up with $n\ge3$ is to just try $n=1,2,3$ $\endgroup$ – user84413 Feb 15 '16 at 20:58
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We can easily use induction. Base case: $n=3$, then $5^3+6^3=341<343=7^3$. Now the inductive hypotheses: let's assume it's true for some $n=k$, then \begin{align} 7^{k+1}&=7\cdot 7^k\\ &\stackrel{IH}{>}7(5^k+6^k)\\ &=7\cdot 5^k+7\cdot 6^k\\ &>5\cdot 5^k+6\cdot 6^k\\ &=5^{k+1}+6^{k+1} \end{align} and so, by the principle of mathematical induction, we're done, thus, the statement $5^k+6^k<7^k$ is true for any $n\geq 3$.

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