Principle of mathematical induction problem Prove the inequality $4^{2n}>15n$
For $n = 1$,
$4^{2\cdot1}=16>15\cdot1$
Let us assume it is true for $n=k$
$4^{2k}>15k$
 A: Inductive step:
$$\begin{align}4^{2(n+1)} &= 4^{2n + 2} \\&= 16\cdot4^{2n}\end{align}$$
But by the induction assumption, we have $4^{2n} > 15n$. Hence,
$$\begin{align}4^{2(n+1)} &> 16\cdot15n \\
&= (15 + 1)\cdot15n \\
&= 15^2n + 15n\\
&= 15^2\left(n-\frac{1}{15}\right) + 15 + 15n\\
&>15 + 15n\\
&= 15(n+1)\end{align}$$
for all positive integers $n$.
A: Induction is made in following steps 


*

*Prove the statement for $n=1$. 

*Assume the statement is true for $n=k$.

*Use, assumption in $2.$ and sometimes the result of $1.$ to prove that the statement is true for $n=k+1$.


Here for step $3.$ start with the LHS $$4^{2(k+1)}$$ and using step $2.$, i.e. that $$4^{2k}>15k$$ is true, try to show that $4^{2(k+1)}$ is larger than $15(k+1)$.
A: (i) for $n=1$, ok because $4^2>15$.
(ii) Suppose that the inequality is true for $n=k$, i.e., $4^{2k}>15k$.
Then,
$$4^{2(k+1)}=4^{2k}\cdot 4^2>15k\cdot 16=15k(15+1)=15^2k+15k>15+15k=15(k+1). $$
(the last inequality is because $15^2k=15\cdot 15\cdot k>15$). 
Then, for (i) and (ii) we concluse for induction. 
A: Define: $\rm f(n) = \dfrac{4^{2n}}{15n} \implies \dfrac{f(n+1)}{f(n)} = \dfrac{4^{2(n+1)}}{15(n+1)} \cdot \dfrac{15n}{4^{2n}} = \dfrac{16n}{n+1} > 1$

 For $\rm n \ge 1: \frac{16n}{n+1} > 1 \Leftrightarrow 16n > n + 1 \Leftrightarrow 15n > 1 \Leftrightarrow n > \frac{1}{15}$ true.

Base Case: $\rm f(1) = \dfrac{16}{15} > 1$
Inductive Step: $\rm f(n) > 1 \implies f(n+1) = \underbrace{\left(\dfrac{f(n+1)}{f(n)}\right)}{} \cdot \underbrace{f(n)}{} > 1$ since both terms in the product are greater than 1 individually.
This is an example of induction using multiplicative telescopy, a powerful technique emphasized by Bill Dubuque. See this and this for similar examples.
