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Thread: Estimate the sum to the nearest whole number

  1. #1
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    Estimate the sum to the nearest whole number

    Estimate the sum to the nearest whole number without the use of a calculator/computer.


    $\displaystyle \displaystyle\sum_{n = 625}^{9999} \sqrt[4]{n + 0.5}$
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  2. #2
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    Re: Estimate the sum to the nearest whole number

    amazing

    eagerly awaiting the simple explanation
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    Re: Estimate the sum to the nearest whole number

    Quote Originally Posted by greg1313 View Post
    Estimate the sum to the nearest whole number without the use of a calculator/computer.


    $\displaystyle \displaystyle\sum_{n = 625}^{9999} \sqrt[4]{n + 0.5}$
    First thoughts. Here's one estimate:

    If $\displaystyle n=5$ then $\displaystyle \sqrt[4]{n+0.5} \approx \sqrt[4]{625} = 5 $

    If $\displaystyle n=9999$ then $\displaystyle \sqrt[4]{n+0.5} \approx \sqrt[4]{10000} = 10 $


    So the first number in the sum is approx 5 and the last number is approx 10. As n increases, the value of $\displaystyle \sqrt[4]{n+0.5}$ also increases.

    There are 9999 - 625 +1 = 9375 terms in the sum.

    Let's assume it is an AP (it isn't, but this will give us an approximation of the sum since the range of values 5 to 10 is quite small and there are a lot of terms).

    So, using the sum of an AP formula, Sum $\displaystyle \approx \frac{9375}{2}(5+10) \approx70313$

    (This gives the same result as if we say each term is approx 7.5 (ie halfway between 5 and 10)).

    So an estimate is 70313.

    Of course there would be other ways and this probably isn't the best!
    Last edited by Debsta; Feb 3rd 2019 at 02:26 PM.
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    Re: Estimate the sum to the nearest whole number

    A (much) better estimate:


    $\displaystyle \int_{625}^{9999} (n+0.5)^\frac{1}{4} dn$

    $\displaystyle \approx \frac{4}{5} (10000^ \frac{5}{4} - 625^\frac{5}{4})$

    $\displaystyle = \frac{4}{5} (10^5 - 5^5)$

    $\displaystyle = 77 500 $
    Last edited by Debsta; Feb 3rd 2019 at 02:49 PM.
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