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Math Help - Need help with math induction, I think I'm on right track

  1. #1
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    Need help with math induction, I think I'm on right track

    2. a) Show by mathematical induction that

    1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - n+2/2^n

    My work:

    1. Show p(1) true. I did this part.

    2. Assume p(k) true => assume p(k+1) true

    I'm a bit stuck here. Here's how I set it up

    (2 - k+2 / 2^k) + (k+1 / 2^k+1) = (2 - k+2 / 2^k) + (k+1 / 2^k+1)

    I think all I have to do is simplify the right side so that it equals 2 - k+3 / 2^k+1. Please help.
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  2. #2
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    Quote Originally Posted by thekrown View Post
    2. a) Show by mathematical induction that

    1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - n+2/2^n
    Does the right-hand side above mean the following?

    2\, -\, \frac{n\, +\, 2}{2^n}

    Thank you!
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  3. #3
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    Yes that is correct. I'm not really sure how to simplify the right to achieve the desired result. The exponents are bugging me.
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  4. #4
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    To do an induction proof, you assume the n = k step, and then try to show the n = k + 1 step. You don't assume the conclusion; you have to prove it.

    I'm not sure what you're doing with the n = k + 1 step...?

    Since the previous (assumption) step was for n = k, you have assumed the following:

    \frac{1}{2^1}\, +\, \frac{2}{2^2}\, +\, \frac{3}{2^3}\, +\, ...\, +\, \frac{k}{2^k}\, =\, 2\, - \frac{k\, +\, 2}{2^k}

    You are now needing to work with:

    \frac{1}{2^1}\, +\, \frac{2}{2^2}\, +\, \frac{3}{2^3}\, +\, ...\, +\, \frac{k}{2^k}\, +\, \frac{k\, +\, 1}{2^{k+1}}

    ...and you hope to manipulate the above to be in the form:

    2\, -\, \frac{k\, +\, 3}{2^{k+1}}

    The first proof step is (usually) to substitute from the assumption step. In this case, we get:

    \left(2\, - \frac{k\, +\, 2}{2^k}\right)\, +\, \frac{k\, +\, 1}{2^{k+1}}

    A good first step might be to convert the second term to a common denominator with the third term, noting that 2^{k+1}\, =\, 2\times 2^k.
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