Inequality 2 to the n greater than n squared

Statement

Show that $2^n>n^2$ for every $n\ge 5$.

Solution

Base ($n=5$): $2^5=32>25=5^2$. True.

Step: suppose $2^k>k^2$ with $k\ge 5$. Then $2^{k+1}=2\cdot 2^k>2k^2.$ It is enough to show that $2k^2\ge(k+1)^2$. Indeed $2k^2-(k+1)^2 = k^2-2k-1 = (k-1)^2-2>0 \quad\text{per } k\ge 5.$ Hence $2^{k+1}>2k^2\ge(k+1)^2$. By induction the inequality holds for every $n\ge 5$. ∎

Links

Topics: Set theory
Concepts: Principle of induction
Methods: Induction
Skills: Proving
Exercise type: Proof