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Thread: Arithmetic progression

  1. #1
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    Arithmetic progression

    What's the sum of all natural numbers from 1 to 99 that dont divide on 3?
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    Re: Arithmetic progression

    Quote Originally Posted by Mustafashama View Post
    What's the sum of all natural numbers from 1 to 99 that dont divide on 3?
    I believe you mean to say, the sum of all natural numbers from 1 to 99 that are not divisible by 3 ...

    $\displaystyle \sum_{n=1}^{99} n - \sum_{n=1}^{33} 3n$
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    Re: Arithmetic progression

    Quote Originally Posted by skeeter View Post
    I believe you mean to say, the sum of all natural numbers from 1 to 99 that are not divisible by 3 ...

    $\displaystyle \sum_{n=1}^{99} n - \sum_{n=1}^{33} 3n$
    The easiest way for a pre-calculus student to do this is to group terms together:

    $\begin{align*}1+2+\cdots +98+99 & =(1+99)+(2+98)+\cdots +(49+51)+50 \\ & = \underbrace{100+100+\cdots +100+100}_{49\text{ times}}+50\\ & = 49\cdot 100 + 50\\ & = 4950\end{align*}$

    Then for all numbers in that range that are divisible by 3:
    $\begin{align*}3+6+\cdots +96+99 & = 3 (1+2+\cdots +32+33) \\ & = 3 ((1+33)+(2+32)+\cdots +(16+18)+17)\\ & = 3 (\underbrace{34+34+\cdots +34}_{16\text{ times}}+17) \\ & = 3 (16\cdot 34 +17)\end{align*}$

    Subtract the two: $4950-1683=3267$
    Last edited by SlipEternal; Jun 8th 2017 at 02:25 AM.
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    Re: Arithmetic progression

    Quote Originally Posted by skeeter View Post
    I believe you mean to say, the sum of all natural numbers from 1 to 99 that are not divisible by 3 ...

    $\displaystyle \sum_{n=1}^{99} n - \sum_{n=1}^{33} 3n$
    The sum of numbers that are non divisible on 3
    And i think u should use the sum arithmetic progression rule
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    Re: Arithmetic progression

    Quote Originally Posted by Mustafashama View Post
    The sum of numbers that are non divisible on 3
    And i think u should use the sum arithmetic progression rule
    I did ...

    $\displaystyle 1+2+3+...+n = S_n=\dfrac{n}{2}[2 + (n-1)] = \dfrac{2n + n(n-1)}{2} = \dfrac{n(n+1)}{2} \implies \sum_{i=1}^n i = \dfrac{n(n+1)}{2}$


    $\displaystyle \sum_{n=1}^{99} n - 3\sum_{n=1}^{33} n = \dfrac{99(100)}{2} - 3 \cdot \dfrac{33(34)}{2} = \dfrac{99(100-34)}{2} = 99 \cdot 33 = 3267$
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