## Thomas' Calculus 13th Edition

By mathematical induction, the formula is true for any integer $n\geq 5.$
$2^{1}\gt 1^{2}$ $2^{2}=2^{2}$ $2^{3}=8\leq 9=3^{2}$ $2^{4}=16=4^{2}$ $2^{5}=32\gt 25=5^{2}$ $2^{6}=64\gt 6^{2}$ $2^{7}=128\gt 7^{2} \quad ...$ So, we want to prove the statement for $n\geq 5$ --- $1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for $n=5$. We just did this. Yes, it is true when $n=5.$ $2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k\geq 5$, THEN it also holds for the next integer, $n=k+1$. Let us assume that for $n=k\geq 5$, $2^{k}\gt k^{2}$ (we call this the step hypothesis) For $n=k+1$, the LHS equals$\quad 2^{k+1}$, which we rewrite as $LHS=(2^{k})\cdot 2\quad$ and we apply the hypothesis $2^{k+1} \gt k^{2}\cdot 2$ now, if $2k^{2}$ is greater than $(k+1)^{2}$, we are done (this would make the formula true for n=k+1). $(k+1)^{2}=k^{2}+2k+1$ now, since $k\geq 5$, it follows that $2\displaystyle \lt\frac{k}{2}$ and $1\displaystyle \lt\frac{k^{2}}{2}$, so: $(k+1)^{2} \lt k^{2}+\displaystyle \frac{k^{2}}{2}+\frac{k^{2}}{2}$ $(k+1)^{2} \lt 2k^{2}$ which makes the formula true for $n=k+1.$ By mathematical induction, the formula is true for any integer $n\geq 5.$