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eliassaab Group Title

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

  • one year ago
  • one year ago

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

    • one year ago
  2. eliassaab Group Title
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    The proof I know does not need induction.

    • one year ago
  3. sauravshakya Group Title
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    @mukushla got any clue?

    • one year ago
  4. mukushla Group Title
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    nope :(

    • one year ago
  5. eliassaab Group Title
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    Hint \[ 4^n=(1+1)^{2n}> {2n \choose n} \]

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

    • one year ago
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