## Precalculus (6th Edition) Blitzer

Let ${{S}_{n}}$ ${{2}^{n}}>{{n}^{2}}$ for $n\ge 5$ Let ${{S}_{n}}$ be a statement involving the positive integer n. If 1. ${{S}_{5}}$ is true, and 2. Let us assume that the ${{S}_{k}}$ be true for some integer $k$. Now, one has to use ${{S}_{k}}$ to prove ${{S}_{k+1}}$ is also true. If one is able to prove ${{S}_{k+1}}$ then by the principle of mathematical induction the statement ${{S}_{n}}$ is true for all positive integers $n\ge 5$. Let ${{S}_{n}}:{{2}^{n}}>{{n}^{2}}$ For, ${{S}_{5}}$ put $n=5$ in the equation; $\left( {{2}^{5}} \right)>\left( {{5}^{2}} \right)$ By simplifying, one gets $32>25$ Now, one has satisfied the first equation. Going to the second condition, assume that $S\left( n \right)$ is true for $n=k$ and prove that the statement is true for $n=k+1$. ${{S}_{k}}:{{2}^{k}}>{{k}^{2}}$ (I) And ${{S}_{k+1}}:{{2}^{k+1}}>{{\left( k+1 \right)}^{2}}$ ${{S}_{k+1}}:{{2}^{k+1}}>{{\left( k+1 \right)}^{2}}$ Simplify the terms \begin{align} & {{2}^{k+1}}=2\left( {{2}^{k}} \right) \\ & >2\left( {{k}^{2}} \right) \\ & ={{k}^{2}}+{{k}^{2}} \end{align} Now, ${{k}^{2}}>2k+1$ for all the values of $k>5$ So, the above equation can be written as \begin{align} & {{2}^{k+1}}=2({{2}^{k}}) \\ & >2\left( {{k}^{2}} \right) \\ & ={{k}^{2}}+{{k}^{2}} \end{align} \begin{align} & >{{k}^{2}}+2k+1 \\ & >{{\left( k+1 \right)}^{2}} \end{align} This inequality for $k+1$ is the same as the previous one which means it has satisfied the second condition of mathematical induction. Therefore, by the method of extended mathematical induction, it is proved that ${{2}^{n}}>{{n}^{2}}$ for $n\ge 5$.