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    The sum of the first n terms of a series is n(n+2). Find the first three terms of the series.


    I don't have a clue! All I've got is sn=n(n+2)
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    (Original post by Lucofthewoods)
    The sum of the first n terms of a series is n(n+2). Find the first three terms of the series.


    I don't have a clue! All I've got is sn=n(n+2)
    Use the formula for the term of the first n terms of a series, and you're given to find the first three terms so when n=3. Put it in your formula to get the answer.
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    (Original post by Lucofthewoods)
    The sum of the first n terms of a series is n(n+2). Find the first three terms of the series.


    I don't have a clue! All I've got is sn=n(n+2)
    Let the nth term of your series be given by u_n and the sum of the first n terms be S_n.
    Note that: S_1=u_1.

    You know a formula for S_n. So if you sub in n=1 you can find u_1.

    What can you say about S_2 and its relation to S_1 and u_2?
    What about S_3?
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    (Original post by Lucofthewoods)
    The sum of the first n terms of a series is n(n+2). Find the first three terms of the series.


    I don't have a clue! All I've got is sn=n(n+2)
    Call your arithmetic progression  a_n .  a_1 = a ,  a_2 = a + d ,  a_3 = a+2d ,  a_n = a + (n-1)d . The partial sums of this progression, aka the sum of the first n terms, is found by adding up the first n terms of our arithmetic progression.  a_1 + a_2 + a_3 + a_4 + ... + a_n this can be compacted by using sigma notation  \sum\limits_{n=1}^n a_n = a_1 + a_2 + a_3 + a_4 + ... + a_n . We are told what the sum of the first  n terms of our progression is  n(n+2) . So,  S_n= \sum\limits_{n=1}^n a_n = n(n+2)

    Notice, that when  n = 1,2,3

     S_1 = \sum\limits_{n=1}^1 a_n = a_1
     S_2 = \sum\limits_{n=1}^2 a_n = a_1 + a_2
     S_3 = \sum\limits_{n=1}^3 a_n = a_1 + a_2 + a_3

    You now need to find,  a_1, a_2 and  a_3 . Good luck, I hope what I've said helps.
 
 
 
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