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Mr.Math

  • 3 years ago

Find the sum \(\large \sum_{k=1}^{n} k!(k^2+k+1).\)

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  1. FoolForMath
    • 3 years ago
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    \( \large \sum \limits_{k=1}^{n} k!(k^2+k+1) = (n+1)!(n+1)-1 \) PS: Latex Tip included ;-)

  2. Mr.Math
    • 3 years ago
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    You should show work, so others can learn :-)

  3. Zarkon
    • 3 years ago
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    \[k^2+k+1=(k+1)^2-k\]

  4. Zarkon
    • 3 years ago
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    is my hint good enough?

  5. Mr.Math
    • 3 years ago
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    Yes.

  6. FoolForMath
    • 3 years ago
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    One more hint, \( \large \sum \limits_{k=1}^{n} k\times k! = (n+1)! -1\)

  7. FoolForMath
    • 3 years ago
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    oops MR.Math is great :)

  8. Mr.Math
    • 3 years ago
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    lol, I already know the answer. I just wanted to post something, and I think it's a nice problem.

  9. FoolForMath
    • 3 years ago
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    haha :)

  10. FoolForMath
    • 3 years ago
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    Mr.Math here: http://openstudy.com/study#/updates/4f2ebf12e4b0571e9cbaf294 ;-)

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