anonymous
  • anonymous
Let \( \pi(n)\) be the number of primes less or equal to n. Show that \[ n^{\pi(2n)-\pi(n)}<4^{n} \]
Meta-math
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
sir i tried induction..but no result are we supposed to prove by induction or not?
anonymous
  • anonymous
The proof I know does not need induction.
anonymous
  • anonymous
@mukushla got any clue?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
nope :(
anonymous
  • anonymous
Hint \[ 4^n=(1+1)^{2n}> {2n \choose n} \]
anonymous
  • anonymous
for \(n\le p_k\le 2n\)\[\prod p_k | \frac{(n+1)(n+2)...(2n)}{n!}={2n \choose n} \Rightarrow\prod p_k <{2n \choose n}\]\[n^{\pi(2n)-\pi(n)}< \prod p_k<\binom{2n}{n}< 2^{2n}=4^{n}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.