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eliassaab

  • 3 years ago

Let \( \pi(n)\) be the number of primes less or equal to n. Show that \[ n^{\pi(2n)-\pi(n)}<4^{n} \]

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  1. mukushla
    • 3 years ago
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    sir i tried induction..but no result are we supposed to prove by induction or not?

  2. eliassaab
    • 3 years ago
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    The proof I know does not need induction.

  3. sauravshakya
    • 3 years ago
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    @mukushla got any clue?

  4. mukushla
    • 3 years ago
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    nope :(

  5. eliassaab
    • 3 years ago
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    Hint \[ 4^n=(1+1)^{2n}> {2n \choose n} \]

  6. mukushla
    • 3 years ago
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    for \(n\le p_k\le 2n\)\[\prod p_k | \frac{(n+1)(n+2)...(2n)}{n!}={2n \choose n} \Rightarrow\prod p_k <{2n \choose n}\]\[n^{\pi(2n)-\pi(n)}< \prod p_k<\binom{2n}{n}< 2^{2n}=4^{n}\]

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