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Prove $5n + 6 \leqslant n^2$ holds for all $n \geqslant N$ by induction. Here $N$ is the answer you get in (a).

For (a) I got $6$ and I proceeded as follows:

Base case: $n = 6$: $5(6)+6 \leqslant 6^22$, $36 \leqslant 36$ therefore base case is true.

Assume $5k + 6 \leqslant k^2$ for $k \geqslant 6$.

Induction: $5(k+1) + 6 \leqslant (k+1)^2$ is true $$5k + 5 + 6 \leqslant k^2 + 2k +1$$ I'm some how confused as to what I need to do next.

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You are almost done. Now note that, since $5k+6\leqslant k^2$, then $5k+5+6=5k+6+5\leqslant k^2+5$ and that $5\leqslant2k+1$, since $k\geqslant6$. So,$$5k+5+6\leqslant k^2+2k+1=(k+1)^2.$$

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