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I am trying to prove the following statement:

For every nonnegative integer $n$, $1+6n \le 7^n$.

I did the base case where $n=0$ but am having trouble manipulating the inductive step. So far I have $7^n+6$. I need that $6$ to be a $7$ so I can have $7^{n+1}$. I am not sure how to make that happen though.

Thanks for the help.

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1 Answer 1

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$7^n+6\leq7^n+7\leq7^n+7^n\leq2\cdot7^n\leq7\cdot7^n=7^{n+1}$ if $n\geq1$

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  • $\begingroup$ Can you keep manipulating that right side simply because its an inequality? 7^n +6 is for sure less than 7^n+7 and so on. Thank you! $\endgroup$
    – Vasi
    May 3, 2015 at 20:50
  • $\begingroup$ @Vasi Well, you can keep manipulating expressions if the inferences you make are true. $\endgroup$
    – Guest
    May 3, 2015 at 20:53
  • $\begingroup$ @Vasi If you want to prove $A\le B$, the most common approach is by doing something like $A\le a\le b\le c\le B$, where all the inequality steps are known to hold, which proves that $A\le B$. $\endgroup$
    – user26486
    May 3, 2015 at 20:53
  • $\begingroup$ Thank you this helps a lot, now must my base case be for n=1 and not n=0? $\endgroup$
    – Vasi
    May 4, 2015 at 17:40
  • $\begingroup$ @Vasi The result is true for $n\geq0$. Hence the base case would be $n=0$. $\endgroup$
    – Guest
    May 4, 2015 at 18:23

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