# Prove result?

• May 23rd 2009, 09:59 PM
fardeen_gen
Prove result?
Given that $\displaystyle n^4 < 10^n$ for a fixed positive integer $\displaystyle n\geq 2$, prove that $\displaystyle (n + 1)^4 < 10^{n + 1}$.
• May 23rd 2009, 11:47 PM
Moo
Hello,
Quote:

Originally Posted by fardeen_gen
Given that $\displaystyle n^4 < 10^n$ for a fixed positive integer $\displaystyle n\geq 2$, prove that $\displaystyle (n + 1)^4 < 10^{n + 1}$.

$\displaystyle (n+1)^4=n^4+4n^3+6n^2+4n+1$

From the given inequality, it follows that $\displaystyle n^3<\tfrac{10^n}{n}$, $\displaystyle n^2<\tfrac{10^n}{n^2}$, $\displaystyle n<\tfrac{10^n}{n^3}$

Hence $\displaystyle (n+1)^4<10^n\left(\frac{4}{n}+\frac{6}{n^2}+\frac{ 4}{n^3}+\frac{1}{10^n}\right)$

Now we're left to prove that $\displaystyle \frac{4}{n}+\frac{6}{n^2}+\frac{4}{n^3}+\frac{1}{1 0^n}\leq 10$

Since $\displaystyle n\geq 2$, it follows that $\displaystyle \frac 1n\leq \frac 12$ and (it's not necessary..) $\displaystyle \frac{1}{10^n}\leq \frac{1}{100}$

Therefore $\displaystyle \frac{4}{n}+\frac{6}{n^2}+\frac{4}{n^3}+\frac{1}{1 0^n}\leq \frac 42+\frac 64+\frac 48+\frac{1}{100}$

And this is obviously less than 10 :)