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.


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,$$5k+5+6\leqslant k^2+2k+1=(k+1)^2.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.