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Math Help - Arithmetic sequence problem

  1. #1
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    Arithmetic sequence problem

    The sum of the first n terms of a certain progression is given by Sn= n^2 +3n
    Find an expression for Sn-1 and deduce an expression for Un
    I managed to figure out Un is 2n+2 but i cant quite get my head around the Sn-1 any help on this would be greatly appreciated
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  2. #2
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    find S1 abd S2.

    S1 = U1 i.e. first term.

    S2 - S1 = U2.

    U2 - U1 = d i.e. the common difference.

    Now proceed.
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  3. #3
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    thx but im still unsure sorry i feel realy thick right now i have the common diff d= 2 the first 4 terms are 4, 6, 8, 10 U1= 4 U2= 6 i still not sure what it wants when it asks for an expression for Sn-1
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  4. #4
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    Quote Originally Posted by pmh118 View Post
    The sum of the first n terms of a certain progression is given by Sn= n^2 +3n
    Find an expression for Sn-1 and deduce an expression for Un
    I managed to figure out Un is 2n+2 but i cant quite get my head around the Sn-1 any help on this would be greatly appreciated
    Hi pmh118,

    S_n=n^2+3n

    S_{n-1}=(n-1)^2+3(n-1)=n^2-2n+1+3n-3=n^2+n-2

    U_1+U_2+U_3+....+U_{n-1}=S_{n-1}

    U_1+U_2+U_3+...+U_{n-1}+U_{n}=S_n

    U_n=S_n-S_{n-1}=n^2+3n-n^2-n+2=2n+2
    Last edited by Archie Meade; October 5th 2010 at 06:03 AM.
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  5. #5
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    Hello, pmh118!

    \text{The sum of the first }n\text{ terms of an arithmetic progression}

    . . \text{is given by: }\;S_n \:=\: n^2 +3n


    \text{(a) Find an expression for }S_{n-1}

    \text{(b) Deduce an expression for }U_n


    \text{I managed to figure out }U_n \:=\:2n+2} . . . . no

    \text{but I can't quite get my head around the }S_{n-1}

    (a) To find S_{n-1}, replace \,n with \,n-1.

    . . . S_{n-1} \;=\;(n-1)^2 + 3(n-1) \;=\;n^2-2n+1+3n-3

    . . . \boxed{S_{n-1} \;=\;n^2 + n -2}



    (b) The common difference \,d is the difference of two consecutive terms.

    . . . d \;=\;S_n - S_{n-1} \;=\;(n^2 + 3n) - (n^2 + n - 2) \;=\;2n+2

    . . . d \:=\:2(n+1) .[1]


    . . . Sum of the first \,n terms: . S_n \;=\;\dfrac{n}{2}\bigg[2a + (n-1)d\bigg]

    . . . \text{We have: }\;\dfrac{n}{2}\bigg[2a + (n-1)d\bigg] \;=\;n^2 +3n

    . . . Substitute [1]: . \dfrac{n}{2}\bigg[2a + (n-1)2(n+1)\bigg] \;=\;n^2+3n

    . . . Solve for \,a\!:\;\;a \;=\;4 + n - n^2 .[2]



    . . . The \,n^{th} term is: . U_n \;=\;a + (n-1)d

    . . . Substitute [1] and [2]: . U_n \;=\;(4+n-n^2) + (n-1)2(n+1)

    . . . . . . . . . . . . . . . . . . . . \boxed{U_n \;=\;n^2+n+2}
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  6. #6
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    Hi Soroban,

    S_n - S_{n-1} is not equal to d, but it is U_n
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  7. #7
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    WoW Thanks all this has really helped me, I think it was the way i was multiplying out of the brackets i was missing the -2n from Sn-1 = (n – 1)2 +3(n – 1) = n2 –2n +1+3n – 3
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