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Math Help - Sequence question, is this correct?

  1. #1
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    Sequence question, is this correct?

    hi everyone

    Question

    a)Calculate the following sequence:
    {(n^3)/(n^3+n^2+1)}^5,n=0


    this is how i did it...
    n = 1, =1/3
    n=2,=8/13
    n=3,27/37
    n=4,64/81
    n=5, 125/151

    ans: 1/3,8/13,27/37,64/81,125/151

    b){an}^6 ,n=0 if a(n+1)= (an+3)/(an+1), a0=2

    a1= (2+3)/(2+1) = 5/3
    a2=(3+3)/(3+1)=3/2
    a3=(4+3)/(4+1)=7/5
    a4=(5+3)/(5+1)=8/6
    a5=(6+3)/(6+1)=9/7
    a6=(7+3)/(7+1)=10/8

    ans:5/3,3/2,7/5,4/3,9/7,5/4


    did i do it right?please help advise.thank you
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Detroit, MI
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    Quote Originally Posted by anderson View Post
    hi everyone

    Question

    a)Calculate the following sequence:
    {(n^3)/(n^3+n^2+1)}^5,n=0

    this is how i did it...
    n = 1, =1/3
    n=2,=8/13
    n=3,27/37
    n=4,64/81
    n=5, 125/151

    ans: 1/3,8/13,27/37,64/81,125/151

    b){an}^6 ,n=0 if a(n+1)= (an+3)/(an+1), a0=2

    a1= (2+3)/(2+1) = 5/3
    a2=(3+3)/(3+1)=3/2
    a3=(4+3)/(4+1)=7/5
    a4=(5+3)/(5+1)=8/6
    a5=(6+3)/(6+1)=9/7
    a6=(7+3)/(7+1)=10/8

    ans:5/3,3/2,7/5,4/3,9/7,5/4


    did i do it right?please help advise.thank you
    The first problem Looks good. Iexamined your method, but did not check every part of your arithmitic, so double check what you have done. Again, your method is correct.

    But, wait. In the secendon problem, a_2=\frac{\frac{5}{3}+3}{\frac{5}{3}+1} .

    a_3=\frac{a_2+3}{a_2+1}\mbox{ and }a_4=\frac{a_3+3}{a_3+1}\mbox{ and }a_5=\frac{a_4+3}{a_4+1}\mbox{ and etc...}
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  3. #3
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    question b is
    { a_n} ^{6}_{n=0} if a_{n+1} = ( a_n+3)/( a_n+1)

    is my answer correct, your working is slightly different.thank you for help & guidance.
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