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Stuck on AS Further Maths Series Question! watch

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    I wrote the series as an expression first which gave me:

    n(n+1)(2n+1)/2 - n(n+1) -1

    I then simplified this expression to n(n+1)(2n-1)/2 - 1.

    I'm unsure on what steps I need to take to get to the answer, which is:
    n(2n+3)(n-1)/2


    See question below:
    Any help would be appreciated!
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    (Original post by dotoswift)
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    That's not entirely correct.

    Note that \displaystyle \sum_{r=1}^n 3r^2-2r-1 = 3\sum_{r=1}^n r^2 - 2 \sum_{r=1}^n r - \sum_{r=1}^n 1

    Now note the last sum - it is not 1
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    (Original post by RDKGames)
    That's not entirely correct.

    Note that \displaystyle \sum_{r=1}^n 3r^2-2r-1 = 3\sum_{r=1}^n r^2 - 2 \sum_{r=1}^n r - \sum_{r=1}^n 1

    Now note the last sum - it is not 1

    Is this the first step then? :

    n(n+1)(2n+1)/2 -n(n+1) - (3-2-1)
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    (Original post by dotoswift)
    Is this the first step then? :

    n(n+1)(2n+1)/2 -n(n+1) - (3-2-1)
    Why? Where did 3-2-1 come from? Try to justify every move you make otherwise you end up writing rubbish without a reason.

    The first two expressions are correct, so don't try to change those.

    Summing up 1 an n number of times means 1+1+1+1+...+1 a total of n times. Hence what is \displaystyle \sum_{r=1}^{n} 1 ?
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    (Original post by RDKGames)
    Why? Where did 3-2-1 come from? Try to justify every move you make otherwise you end up writing rubbish without a reason.

    The first two expressions are correct, so don't try to change those.

    Summing up 1 an n number of times means 1+1+1+1+...+1 a total of n times. Hence what is \displaystyle \sum_{r=1}^{n} 1 ?
    It is 1n.
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    (Original post by dotoswift)
    It is 1n.
    Yes, so then proceed with that correction. Factor out n and simplify the bracket.
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    (Original post by RDKGames)
    Yes, so then proceed with that correction. Factor out n and simplify the bracket.
    Thanks so much for your help.

    I finally got the answer!
 
 
 
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