Problem Statement
Use the principle of mathematical induction to prove that:
$$2^n > n^2$$
for all integers $n \ge 5$.
Verified Solution & Marking Scheme
Base Step n = 5
For $n = 5$:
$\text{LHS} = 2^5 = 32$.
$\text{RHS} = 5^2 = 25$.
Since $32 > 25$, the statement is true for the base case $n = 5$.
Inductive Hypothesis
Assume the statement is true for some positive integer $k \ge 5$, that is:
$$2^k > k^2$$
Inductive Step (Prove for n = k + 1)
We need to prove that $2^{k+1} > (k + 1)^2 = k^2 + 2k + 1$.
Starting with the LHS:
$$2^{k+1} = 2 \times 2^k$$
Using the inductive hypothesis $2^k > k^2$:
$$2^{k+1} > 2k^2 = k^2 + k^2$$
We now compare $k^2$ with $2k + 1$ for $k \ge 5$:
$$k^2 - 2k - 1 = (k - 1)^2 - 2$$
For $k \ge 5$, $k - 1 \ge 4 \implies (k - 1)^2 \ge 16 > 2$.
Hence $k^2 > 2k + 1$.
Therefore:
$$2^{k+1} > k^2 + k^2 > k^2 + 2k + 1 = (k + 1)^2$$
Formal IB Conclusion
Since the statement is true for $n = 5$, and whenever it is true for $n = k$ it is also true for $n = k + 1$, by the principle of mathematical induction, $2^n > n^2$ is true for all integers $n \ge 5$.