Prove that $3^{2015}-2^{2015}>2016^2$

Question

I have to prove:$$3^{2015}-2^{2015}>2016^2$$ by induction.

I wanted to use the standard factorizing formula: $$a^{mn}−b^{mn}=(a^n−b^n)(a^{n(m−1)}+a^{n(m−2)}b^n+...+a^nb^{n(m−2)}+b^{n(m−1)}).$$ $2015=403\cdot 5$

What would be the next step?

Answer 1

We will prove for a more general case , i.e :

$$3^x - 2^x > (x+1)^2 \quad\quad \text{ For x > 3 }$$

For base case , $3^3 - 2^3 = 19 > 16$

Now let be true for a value $n$ , such that $3^n - 2^n > (n+1)^2$

$$\begin{align}3^n - 2^n + 2n+3& > (n+1)^2 + 2n+3\\ 3^n - 2^n + 2n+3& > (n+2)^2 \end{align}$$ Now it is left to show that $3^{n+1} - 2^{n+1} >3^n - 2^n + 2n+3 $.We show it by assuming it is true.

$$\begin{align}3^{n+1} - 2^{n+1} &>3^n - 2^n + 2n+3 \\ 3^{n+1} - 3^n + 2^n - 2^{n+1} &> 2n+3 \\ 2.3^n - 2^n &> 2n+3\end{align}$$

Which follows from the fact that $3^n - 2^n > (n+1)^2 > 2n+3$.

Hence our induction is complete.

Answer 2

By the binomial theorem, $$ 3^{2015}=(2+1)^{2015} > 2^{2015}+\binom{2015}{1}2^{2014} $$ The result follows because $2^{2014} > 2^{30} > 10^{9} > 10^7 > 9 \cdot 10^6 = 3000^2 > 2016^2$.

Answer 3

We have \begin{multline*}3^{2015}-2^{2015}=2^{2014}+2^{2013}\cdot 3+\dots+3^{2014}>2^{2014}+2^{2013}+\dots+1=2^{2015}-1\\>2016^2=4064256.\end{multline*}