Prove that 2n>n2 for all natural numbers n≥5.

Let P(n) be the proposition that 2n>n2.

Verify P(5):

We have 25=32 and 52=25; hence, 32>25 and P(5) is true.

Assume P(k) is true:

Show that P(k+1) is true:

To complete the induction, we need to show that

Since (k+1)2=k2+2k+1, it suffices to prove

The quadratic k2−2k−1 is concave up and has roots k=22±√8​=1±√2. So for k≥5>1+√2, k2−2k−1>0⇒2k2>(k+1)2.

Thus, 2k+1>(k+1)2, and P(k+1) is true.

Conclusion:

Since P(5) is true and if P(k) then P(k+1) is true, by induction 2n>n2 for all n≥5.