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Math Help - consecutive numbers

  1. #1
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    consecutive numbers

    hello everyone!

    the square of the mean of the consecutive integers 1,2,3,4,5 is 9
    the mean of the squares of the same 5 numbers is 11

    show algebraically that the square of any five consecutive integers is always 2 less than the mean of the squares of those 5 consecutive integers

    Here's what I got, but I have a sneaky feeling I have done something wrong cos I ended up with a quadratic!!
    Attached Thumbnails Attached Thumbnails consecutive numbers-consecutive-numbers-algebra.jpg  
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  2. #2
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    I forgot to include the -2 on the right hand side in the image but I did include it in the working to get the final equation.
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  3. #3
    Moo
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    Hello,

    You also forgot that you had to compare the square of the mean, not the mean.

    And you can't simplify the 5. I'll give you a (BIG ) hint :

    n+n+1+\dots+n+4=5n+10

    The text talks about the square of the mean.

    So it's (\frac{5n+10}{5})^2=(n+2)^2=\dots


    Now the other side.
    The sum of the squares is 5n^2+{\color{red}20}n+30

    Hence, since the text talks about the mean of the squares, it's \frac{5n^2+20n+30}{5}=\dots

    Can you continue ?
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  4. #4
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    OK. Thanks for the reply. I had another go, I'm afraid it still doesn't look right!
    I am enjoying it though!

    OH, that should be n = n^2 at the end, sorry (again!)
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  5. #5
    Moo
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    You've got n=n, which is true.
    Hence, your affirmation is ok.


    But it's better doing this way :

    The square of the mean is :
    (n+2)^2=n^2+4n+4

    The mean of the squares is :
    \frac{5n^2+20n+30}{5}=n^2+4n+6

    And it's obvious that the difference is 2

    Why is it better this way ? Because you're not really supposed to know the result
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