Principle of mathematical induction problem

Question

Prove the inequality $4^{2n}>15n$

For $n = 1$,

$4^{2\cdot1}=16>15\cdot1$

Let us assume it is true for $n=k$

$4^{2k}>15k$

Answer 1

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 $n$.

Answer 2

Induction is made in following steps

1. Prove the statement for $n=1$.
2. Assume the statement is true for $n=k$.
3. 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 $15(k+1)$.

Answer 3

(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.

Answer 4

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.