Prove result?

Hello,

Quote:

Originally Posted by fardeen_gen

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

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

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

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

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

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

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

And this is obviously less than 10 :)