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Unsolved Equation

  1. Sep 27, 2004 #1
    I am kinda new here but heres a problem for you guys:

    There is an equation of the form:
    ((1)^k)+((2)^k)+......+((n-1)^k)+((n)^k) = ((n+1)^k)

    This equation is such that all the numbers starting from1 till n are raised to the power of k and added and the result is (n+1)^k. What should n and k be?

    Apart from ((1)^2) + ((2)^2) = ((3^2)) There isnt any other obvious answer. ( These were the first numbers I tried when I began trying to solve this problem.

  2. jcsd
  3. Sep 27, 2004 #2
    ((1)^2) + ((2)^2) = ((3^2))?????

    Is this true? :confused:
  4. Sep 27, 2004 #3


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    Clearly, it is not.

    But 1+2=3 is true. And there's a more trivial solution : 1^0 = 2^0.
  5. Sep 27, 2004 #4
    It was a typo ...it should have been

  6. Sep 27, 2004 #5
    I just did a mistake.. it was supposed to be

    ((1)^1) + ((2)^1) = ((3)^1)

    I will be careful in the future
  7. Sep 30, 2004 #6
    I belive that if k was an odd number and n was negative then it would be possible to solve it another way. With a negative number in there you can counteract all of the adding of things.
  8. Oct 3, 2004 #7
    is the original problem like this:

    sigma (n^c) where c is any real constant

    i was trying to figure that out, but if yours is

    sigma (c^n) where c is any real constant,

    then i think the sum is {[c^(k+1)]/(c-1)}-[1/(c-1)]

    hope this helps
  9. Nov 8, 2004 #8
    I think about the closest series to that is 1+2+4+8++2^N =2^(n+1)-1. This comes about because [tex] 1 +r +r^2+r^n=\frac{-1+r^_(n+1)}{r-1}[/tex]
    but, of course, 2-1=1, so the denominator disappears.
    Last edited: Nov 8, 2004
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