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Igma notation 2i-1 = 2n, for all n is an element of N

  1. Dec 5, 2008 #1
    prove or disprove

    sigma notation 2i-1 = 2n, for all n is an element of N.

    N = natural numbers
  2. jcsd
  3. Dec 5, 2008 #2
    Re: Induction

    Rewritten using LaTeX,
    [tex]\sum_{i=1}^{n} 2^{i - 1} = 2^n.[/tex]

    Have you made any real attempt at this problem? Do you have a guess as to whether it's true or false? What I like to do on these types of "prove or disprove" questions is just plug in certain (small) values of n, and see if it works. If they all work, perhaps it really is true, and you should try to prove it; otherwise, it's false as you found a counterexample.

    And once you find that it's false, can you guess a correct formula for the sum and prove that?
  4. Dec 8, 2008 #3
    Re: Induction

    Here is my attempt at this problem:

    sigma notation 2i-1= 2n, for all n is an element of N


    sigma notation 21-1= 20 = 1 not equal to 21

    n=1 is an element of N

    sigma notation 2i-1=2n doesn't hold for all n is an element of N

    did i do this right?
  5. Dec 8, 2008 #4
    Re: Induction

    Your solution is correct, although it's slightly unclear, and you should use more words to describe what you're doing.

    You were trying to disprove
    [tex]\sum_{i=1}^{n} 2^{i - 1} = 2^n \text{ for all } n \in \mathbb{N}.[/tex]
    (It's much prettier in LaTeX; you should learn (at least by example) how to use it. [tex]n \in \mathbb{N}[/tex] is read "n in N" or even "natural numbers n" in this case. Click the images to see the code used to make them.) You put in n = 1 and showed that the two sides aren't equal; that is a counterexample, so your disproof is correct.
  6. Dec 8, 2008 #5


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    Science Advisor

    Re: Induction

    Yes, you did! Now what is your answer to the question?

    And you might like to look at
    [tex]\sum_{i= 1}^n 2^{i-1}= 2^n- 1[/itex]
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