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Finding the nth term of a sequence

  1. Dec 10, 2014 #1
    • Member warned about not using the homework template
    Find the nth term of the sequence; ## 2, 5, 10, 17,26..........................## this is how i did it:

    2, 5, 10, 17, 26 can form a difference of 3,5, 7,9.........
    whose nth term is 1 + 2n. now how can apply this to the original sequence to get the nth term?

     
  2. jcsd
  3. Dec 10, 2014 #2

    haruspex

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    Starting with the 2, you add numbers of the form 2r+1. Do you know how to write a summation expression for a sum of numbers of that form?
     
  4. Dec 10, 2014 #3
    thanks i understand the 2r+1 which is the 1+2n which i indicated, my interest here is how to get the nth term yes sum ##Sn= n/2(2a+(n-1)d)## how do i use this?
     
  5. Dec 10, 2014 #4

    haruspex

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    Find the values of a and d which generate the given sequence.
     
  6. Dec 10, 2014 #5
    the nth term is ##n^2+1## now my problem is i know ##a=2## what about d? i have tried using ##d= 2n+1## and i am getting ##Sn= n/2(2n^2-2+3)##
    now how do i move from here to the required ##n^2+1##?
     
  7. Dec 10, 2014 #6

    haruspex

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    No, a and d are constants.
    Write the first two terms in the sequence in terms of a and d. That gives you two equations, two unknowns.
     
  8. Dec 11, 2014 #7
    please assist me i still dont understand..............
     
  9. Dec 11, 2014 #8

    haruspex

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    You have an expression Sn=n/2(2a+(n−1)d) which you believe will generate the sequence given the right values of a and d. In particular, it will need to give the right values for S1 and S2. You know what S1 and S2 are, so that gives you two equations.
     
  10. Dec 11, 2014 #9
    for S1=n/2(2a+(n-1)d gives
    2= 1/2(2.a+(1-1)d) gives 4=2a and S2=2+5=7, 7=2/2(2a+(2-1)d gives 7=2a+d sir, i still dont understand just make it clear for me i am conversant with the general knowledge on arithmetic and geometric.....just give the needed equations where a=2 and d=3 how do you move from here to n^2+1?
     
  11. Dec 11, 2014 #10

    Ray Vickson

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    You were asked to give a formula for the nth term of the sequence; you have done that, so you have answered the question in full. Why do you think you need to "move from here to n^2+1"?
     
  12. Dec 11, 2014 #11

    haruspex

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    So use those to replace a and d in your expression n/2(2a+(n−1)d).
     
  13. Dec 16, 2014 #12
    Thanks i got it....................##(1+1)+3+5+7+9+.##............
    = ##1+(1+3+5+7+9+.............)## forms the series ##1+ n^2##
     
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