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Math Help - Power of intergers

  1. #1
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    Power of intergers

    Find the sum, using the formulas for the sums of the powers of integers.

    11
    (-2-n^3+2n)
    n=1
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by flexus View Post
    Find the sum, using the formulas for the sums of the powers of integers.

    11
    (-2-n^3+2n)
    n=1
    You need to know that:

    \sum_{i=1}^{n}k=kn

    \sum_{i=1}^{n}i=\frac{n(n+1)}{2}

    \sum_{i=1}^{n}i^3=\left(\frac{n(n+1)}{2}\right)^2

    Can you tackle the problem now?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    You need to know that:

    \sum_{i=1}^{n}k=kn

    \sum_{i=1}^{n}i=\frac{n(n+1)}{2}

    \sum_{i=1}^{n}i^3=\left(\frac{n(n+1)}{2}\right)^2

    Can you tackle the problem now?
    the answer is 8712?????
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    i dont think that is the right answer the pride.. i see what you did though. you plugged 11 into n. you used the second equation. thanks for helping. the answer is will not help me if i don't understand this. can someone confirm his answer with a little explanation?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by flexus View Post
    i dont think that is the right answer the pride.. i see what you did though. you plugged 11 into n. you used the second equation. thanks for helping. the answer is will not help me if i don't understand this. can someone confirm his answer with a little explanation?
    Note that \sum_{n=1}^{11}\left(2-n^3+2n\right)=\sum_{n=1}^{11}2-\sum_{n=1}^{11}n^3+2\sum_{n=1}^{11}n.

    Now substitute n=11 into the formulas I listed above and that will give you the answer you're looking for.
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    Quote Originally Posted by Chris L T521 View Post
    You need to know that:

    \sum_{i=1}^{n}k=kn

    \sum_{i=1}^{n}i=\frac{n(n+1)}{2}

    \sum_{i=1}^{n}i^3=\left(\frac{n(n+1)}{2}\right)^2

    Can you tackle the problem now?
    hey im looking at the first equation you gave me and i dont seem to know what to substitute k for?? so n=11 what does k =?
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by flexus View Post
    hey im looking at the first equation you gave me and i dont seem to know what to substitute k for?? so n=11 what does k =?
    K is any constant. In your particular problem, k=2.
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    Quote Originally Posted by Chris L T521 View Post
    K is any constant. In your particular problem, k=2.
    you mean -2 from the oringinal problem correct?

    so if i set it up it would look like this?

    (-2)(11)-11(11+1)/2+(11(11+1)^2/2= 704

    correct?
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  9. #9
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by flexus View Post
    you mean -2 from the oringinal problem correct?

    so if i set it up it would look like this?

    (-2)(11)-11(11+1)/2+(11(11+1))^2/4= 704 (change the value on the RHS and then you should have the answer)

    correct?
    Yes, I meant to say -2 -- my mistake.

    You're so close -- note my corrections in red.
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  10. #10
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    Quote Originally Posted by Chris L T521 View Post
    Yes, I meant to say -2 -- my mistake.

    You're so close -- note my corrections in red.
    omg i wrote the problem wrong. ok let me do this over.

    the correct problem is:

    15
    (-2-n^3+2n)
    n=1

    ok so now i plug that into the three equations.

    (-2)(15)-(15(15+1))/2+15(15+1)^2)/2 the i get:

    -30-240/2+1920 which equals =1770 but this does not match the answer in my book. i must have done something wrong? can you confirm please?
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  11. #11
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by flexus View Post
    omg i wrote the problem wrong. ok let me do this over.

    the correct problem is:

    15
    (-2-n^3+2n)
    n=1

    ok so now i plug that into the three equations.

    (-2)(15)-(15(15+1))/2+15(15+1)^2)/2 the i get:

    -30-240/2+1920 which equals =1770 but this does not match the answer in my book. i must have done something wrong? can you confirm please?
    It should be (-2)(15)- [(15(15+1))/2]^2+2[(15)(15+1)/2]...

    Do you see why?
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  12. #12
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    Quote Originally Posted by Chris L T521 View Post
    It should be (-2)(15)- [(15(15+1))/2]^2+2[(15)(15+1)/2]...

    Do you see why?
    so you are gonna distribute the negative to the right?
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  13. #13
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by flexus View Post
    so you are gonna distribute the negative to the right?
    For the second term? Yes.
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  14. #14
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    Quote Originally Posted by Chris L T521 View Post
    For the second term? Yes.
    ok im gonna try this again. my head is about to explode lol

    (-2)(15)-[(15(15+1))/2]^2 + 2[(15)(15+1)/2]

    = -30 -14400+14400

    =-30

    OK that's what i got. can you confirm please.
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  15. #15
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by flexus View Post
    ok im gonna try this again. my head is about to explode lol

    (-2)(15)-[(15(15+1))/2]^2 + 2[(15)(15+1)/2]

    = -30 -14400+14400

    =-30

    OK that's what i got. can you confirm please.
    The term in red should be 240 (since its not being squared).

    You should end up with -14190
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