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Math Help - Find the 1000th term

  1. #1
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    Find the 1000th term

    1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5.............,k...... is a sequence where the number k appears k times successively. (k=1, 2, 3, 4, 5,...) Find the 1000th term of the sequence.

    S_k=1000

    1+2+3+4+5+...+n=1000

    \frac{n}{2}[2+(n-1)]

    \frac{n}{2}[n+1]=1000

    n(n+1)=2000

    n^2+n-2000=0

    n=-45.22(reject) or 44.22


    should i accept n=44 or n=45 as the answer?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Punch View Post
    1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5.............,k...... is a sequence where the number k appears k times successively. (k=1, 2, 3, 4, 5,...) Find the 1000th term of the sequence.

    S_k=1000

    1+2+3+4+5+...+n=1000

    \frac{n}{2}[2+(n-1)]

    \frac{n}{2}[n+1]=1000

    n(n+1)=2000

    n^2+n-2000=0

    n=-45.22(reject) or 44.22


    should i accept n=44 or n=45 as the answer?

    I don't understood what you done, bur here is my hint:

    Observe when 1,2,3,4,5,... appears in your sequence.

    Try to find a close formula to where k first time appears
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    I don't understood what you done, bur here is my hint:

    Observe when 1,2,3,4,5,... appears in your sequence.

    Try to find a close formula to where k first time appears
    Notice that the question says a sequence where the number k appears k times successively.

    So the sequence k=1,2,3,4,5,...... and so on.

    Sequence k is an arithmetic progression.

    Since the number k appears k times, the sum of the sequence up to the kth term is 1000.

    Hence, 1+2+3+4+5+...+n=1000

    then by using the sum of an arithmetic progression, i used it to find n
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Here is close formula for your sequence:

    f(n)=\left \lfloor {\frac{1+\sqrt{1+8n}}{2}} \right \rfloor
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  5. #5
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    Quote Originally Posted by Punch View Post
    1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5.............,k...... is a sequence where the number k appears k times successively. (k=1, 2, 3, 4, 5,...) Find the 1000th term of the sequence.

    S_k=1000

    1+2+3+4+5+...+n=1000

    \frac{n}{2}[2+(n-1)]

    \frac{n}{2}[n+1]=1000

    n(n+1)=2000

    n^2+n-2000=0

    n=-45.22(reject) or 44.22


    should i accept n=44 or n=45 as the answer?
    Suppose the 1000th term is 44 count the maximum number of terms less than or equal to 44...

    The largest number of terms less than or equal to 44 is:

    1+2+ ... +44=990,

    a contradiction, now try a similar argument with 45.

    CB
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    Suppose the 1000th term is 44 count the maximum number of terms less than or equal to 44...

    The largest number of terms less than or equal to 44 is:

    1+2+ ... +44=990,

    a contradiction, now try a similar argument with 45.

    CB
    Oh yes, i could have used the sum formula
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  7. #7
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    Quote Originally Posted by Punch View Post
    1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5.............,k...... is a sequence where the number k appears k times successively. (k=1, 2, 3, 4, 5,...) Find the 1000th term of the sequence.

    S_k=1000

    1+2+3+4+5+...+n=1000
    What does this have to do with your problem?

    \frac{n}{2}[2+(n-1)]

    \frac{n}{2}[n+1]=1000

    n(n+1)=2000

    n^2+n-2000=0

    n=-45.22(reject) or 44.22


    should i accept n=44 or n=45 as the answer?
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  8. #8
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    Quote Originally Posted by Punch View Post
    1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5.............,k...... is a sequence where the number k appears k times successively. (k=1, 2, 3, 4, 5,...) Find the 1000th term of the sequence.

    S_k=1000

    1+2+3+4+5+...+n=1000

    \frac{n}{2}[2+(n-1)]

    \frac{n}{2}[n+1]=1000

    n(n+1)=2000

    n^2+n-2000=0

    n=-45.22(reject) or 44.22


    should i accept n=44 or n=45 as the answer?
    You are taking the last of the repeated numbers and calling it Tn.

    T1=1
    T3=2, and 1+2=3
    T6=3, and 1+2+3=6
    T10=4, and 1+2+3+4=10

    Hence you want T1000=1+2+3+4+....+n.

    However, you do not know that T1000 is the last of a repeated list of the same number.

    Following CaptainBlack's advice,
    T990=44
    generated from the sum of terms, so it's the last 44.
    Hence T991 is 45 and how many of those follow T990 ?
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  9. #9
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    Quote Originally Posted by Archie Meade View Post
    You are taking the last of the repeated numbers and calling it Tn.

    T1=1
    T3=2, and 1+2=3
    T6=3, and 1+2+3=6
    T10=4, and 1+2+3+4=10

    Hence you want T1000=1+2+3+4+....+n.

    However, you do not know that T1000 is the last of a repeated list of the same number.

    Following CaptainBlack's advice,
    T990=44
    generated from the sum of terms, so it's the last 44.
    Hence T991 is 45 and how many of those follow T990 ?
    I would say 0.22(45) follow T990
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  10. #10
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    Your sequence has a single 1, 2 twice, 3 three times, 4 four times, 5 five times, 6 six times,
    44 forty four times, etc.
    So just go back to CaptainBlack's post and realise what you were calculating with your sum
    (explained in my previous post).
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