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Math Help - Need proof by induction

  1. #1
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    Need proof by induction

    (original question) Prove that the Summation of j=0 to n of (-1/2)^j = (2n+1 + (-1)n) / (3*2n)


    I have a question that includes the following
    P(k) (-1/2)0 + (-1/2)1 + (-1/2)2 + ... + (-1/2)k = (2k+1 + (-1)k) / (3*2k)

    So I have k+1 as

    P(k+1) ((2K+1 + (-1)k) / (3*2k)) + (-1/2)k+1 = (2k+2 +(-1)k+1) / (3*2k+1)
    But I dont know how to get those sides to be equivalent. Can anyone please help?
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  2. #2
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    Re: Need proof by induction

    Quote Originally Posted by Brent1986 View Post
    (original question) Prove that the Summation of j=0 to n of (-1/2)^j = (2n+1 + (-1)n) / (3*2n)


    I have a question that includes the following
    P(k) (-1/2)0 + (-1/2)1 + (-1/2)2 + ... + (-1/2)k = (2k+1 + (-1)k) / (3*2k)

    So I have k+1 as

    P(k+1) ((2K+1 + (-1)k) / (3*2k)) + (-1/2)k+1 = (2k+2 +(-1)k+1) / (3*2k+1)
    But I dont know how to get those sides to be equivalent. Can anyone please help?
    \\\sum\limits_{j = 0}^{N + 1} {{{\left( {\frac{{ - 1}}{2}} \right)}^j}}  = \\\sum\limits_{j = 0}^N {{{\left( {\frac{{ - 1}}{2}} \right)}^j}}  + {\left( {\frac{{ - 1}}{2}} \right)^{N + 1}} =\\ \frac{{{2^{N + 1}} + {{( - 1)}^N}}}{{3 \cdot {2^N}}} + {\left( {\frac{{ - 1}}{2}} \right)^{N + 1}} =\\ \frac{{{2^{N + 2}} + 2{{( - 1)}^N} + {{3( - 1)}^{N+1}}}}{{3 \cdot {2^{N + 1}}}}\\=??
    Last edited by Plato; September 8th 2013 at 02:18 PM.
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  3. #3
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    Re: Need proof by induction

    Im sorry, im feeling kinda dumb can you tell me how I add the 2 equations you have on that 3rd line?
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  4. #4
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    Re: Need proof by induction

    Quote Originally Posted by Brent1986 View Post
    Im sorry, im feeling kinda dumb can you tell me how I add the 2 equations you have on that 3rd line?
    Notice that \frac{{{2^{N + 1}} + {{( - 1)}^N}}}{{3 \cdot {2^N}}} + {\left( {\frac{{ - 1}}{2}} \right)^{N + 1}} = \frac{{{2^{N + 1}} + {{( - 1)}^N}}}{{3 \cdot {2^N}}} + \frac{{{{( - 1)}^{N + 1}}}}{{{2^{N + 1}}}}

    So the \text{LCD is }3\cdot 2^{N+1} thus \frac{{{2^{N + 1}} + {{( - 1)}^N}}}{{3 \cdot {2^N}}} + \frac{{{{( - 1)}^{N + 1}}}}{{{2^{N + 1}}}} = \frac{{{2^{N + 2}} + 2{{( - 1)}^N} + 3{{( - 1)}^{N + 1}}}}{{3 \cdot {2^{N + 1}}}}

    Now notice that 2{( - 1)^N} + 3{( - 1)^{N + 1}} = {( - 1)^{N + 1}}\left[ { - 2 + 3} \right].

    BTW: I do not usually just hand out solutions.
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  5. #5
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    Re: Need proof by induction

    That last part looks similar to the distributive property but our 'a' isn't the same (N vs. N +1) and the b is inverted. Is there a name for that property? Or is it just something I should remember.

    I really appreciate the help, not looking for solutions just understanding!
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  6. #6
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    Re: Need proof by induction

    Quote Originally Posted by Brent1986 View Post
    That last part looks similar to the distributive property but our 'a' isn't the same (N vs. N +1) and the b is inverted. Is there a name for that property? Or is it just something I should remember.
    There are several ways to do this:
    2{( - 1)^N} + 3{( - 1)^{N + 1}} = {( - 1)^N}\left[ {2 - 3} \right] = {( - 1)^N}\left[ { - 1} \right] = {( - 1)^{N + 1}}
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  7. #7
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    Re: Need proof by induction

    Much thanks to you my friend! I spent a few hours today trying to wrap my brain around these questions. I feel like I have a much better understanding now! You've really helped me!
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