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Math Help - Prove result?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove result?

    Given that n^4 < 10^n for a fixed positive integer n\geq 2, prove that (n + 1)^4 < 10^{n + 1}.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by fardeen_gen View Post
    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
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