## anonymous 3 years ago Let $$\pi(n)$$ be the number of primes less or equal to n. Show that $n^{\pi(2n)-\pi(n)}<4^{n}$

1. anonymous

sir i tried induction..but no result are we supposed to prove by induction or not?

2. anonymous

The proof I know does not need induction.

3. anonymous

@mukushla got any clue?

4. anonymous

nope :(

5. anonymous

Hint $4^n=(1+1)^{2n}> {2n \choose n}$

6. anonymous

for $$n\le p_k\le 2n$$$\prod p_k | \frac{(n+1)(n+2)...(2n)}{n!}={2n \choose n} \Rightarrow\prod p_k <{2n \choose n}$$n^{\pi(2n)-\pi(n)}< \prod p_k<\binom{2n}{n}< 2^{2n}=4^{n}$