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Math Help - sequence problem

  1. #1
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    sequence problem

    1+(K+1)(14)+12(1+2+3+....(K-1)+K)=X

    Solve for K?


    Is this solvable? as I said before haven't done maths for years and this is making my head spin. If any one could help that would be great!
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  2. #2
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    Quote Originally Posted by Not2l8 View Post
    1+(K+1)(14)+12(1+2+3+....(K-1)+K)=X

    Solve for K?


    Is this solvable? as I said before haven't done maths for years and this is making my head spin. If any one could help that would be great!
    Sorry I see no obvious pattern.
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  3. #3
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    Hello, Not2l8!

    What a strangely-written problem!
    And is there a typo?


    Solve for k\!:\;\;1 + 14(k+1) + 12(1+2+3+\hdots k) \;=\; {\color{red}k}

    We have: . 1 + 14k + 14 + 12\cdot\frac{k(k+1)}{2} \;=\;k \quad\Rightarrow\quad 1 + 14k + 14 + 6k^2 + 6k \;=\;k

    . . 6k^2 + 19k + 15 \:=\:0 \quad\Rightarrow\quad (2k + 3)(3k + 5) \:=\:0


    Therefore: . k \;=\;-\frac{3}{2},\:-\frac{5}{3}

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  4. #4
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    Hi Soroban, thanks very much for your response.

    Sorry I'm not really a maths guy hence the strangely written problem.
    I think your answer was the one I was looking for but I was trying to solve for X at the end not K.

    1+14(K+1)+12(1+2+3....+K)=X

    And I think by your result I can see

    6k^2+20K+15=X

    which is awesome! That's what I was looking for thanks!
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