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Math Help - Prove by Induction

  1. #1
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    Post Prove by Induction

    How to do this following "Induction sum"?

    n
    SUM (-1/2)^j = (2^n+1 + (-1)^n)/ 3. 2^n
    j=0

    How do we remove the "summation" sign here? Also, how to go about proving it by using "induction"?

    Thanks for you reply!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello

    To prove this by induction, check that the equality is true for n=0. (easy )

    Then, assuming the equality is true for an integer n,

    \begin{aligned}<br />
S_{n+1}=\sum_{j=0}^{n+1}\left(\frac{-1}{2}\right)^j&=\sum_{j=0}^n\left(\frac{-1}{2}\right)^j+\left(\frac{-1}{2}\right)^{n+1}\\<br />
&=\frac{2^{n+1}+(-1)^n}{3\cdot 2^n}+\left(\frac{-1}{2}\right)^{n+1}\\<br />
\end{aligned}

    Multiply the first fraction by \frac{2}{2} and the second one by \frac{3}{3} so that they can be added.

    S_{n+1}=\frac{2}{2}\cdot\frac{2^{n+1}+(-1)^n}{3\cdot 2^n}+\frac{3}{3}\cdot\left(\frac{-1}{2}\right)^{n+1}=\frac{2^{n+2}+2\cdot(-1)^n}{3\cdot2^{n+1}}+\frac{3\cdot(-1)^{n+1}}{3\cdot 2^{n+1}}=\ldots

    Does it help ?
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  3. #3
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    Hi, Thanks for you quick reply man!

    Anyways, I didn't got how did you "expand"

    n
    sum (-1/2)^j into (2^n+1+(-1)^n)/3.2^n etc...
    j =0

    Also, how to get the symbols like you are using? Do we have to write the code for that?

    Thanks
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by robocop_911 View Post
    Hi, Thanks for you quick reply man!

    Anyways, I didn't got how did you "expand"

    n
    sum (-1/2)^j into (2^n+1+(-1)^n)/3.2^n etc...
    j =0
    Quote Originally Posted by flyingsquirrel View Post
    Then, assuming the equality is true for an integer n,
    I assumed the equality \sum_{j=0}^n\left(\frac{-1}{2}\right)^j=\frac{2^{n+1}+(-1)^n}{3\cdot 2^n} was true for n.
    Also, how to get the symbols like you are using? Do we have to write the code for that?
    Yes, you have to write the code for each formula. See this thread.
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  5. #5
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    Quote Originally Posted by flyingsquirrel View Post
    Hello

    To prove this by induction, check that the equality is true for n=0. (easy )

    Then, assuming the equality is true for an integer n,

    \begin{aligned}<br />
S_{n+1}=\sum_{j=0}^{n+1}\left(\frac{-1}{2}\right)^j&=\sum_{j=0}^n\left(\frac{-1}{2}\right)^j+\left(\frac{-1}{2}\right)^{n+1}\\<br />
&=\frac{2^{n+1}+(-1)^n}{3\cdot 2^n}+\left(\frac{-1}{2}\right)^{n+1}\\<br />
\end{aligned}

    Multiply the first fraction by \frac{2}{2} and the second one by \frac{3}{3} so that they can be added.

    S_{n+1}=\frac{2}{2}\cdot\frac{2^{n+1}+(-1)^n}{3\cdot 2^n}+\frac{3}{3}\cdot\left(\frac{-1}{2}\right)^{n+1}=\frac{2^{n+2}+2\cdot(-1)^n}{3\cdot2^{n+1}}+\frac{3\cdot(-1)^{n+1}}{3\cdot 2^{n+1}}=\ldots

    Does it help ?
    Can you expand this further more... I don't get it why are you multiplying and dividing by 2/2 and 3/3.
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by robocop_911 View Post
    Can you expand this further more... I don't get it why are you multiplying and dividing by 2/2 and 3/3.
    I did this so that the two fractions share the same denominator.

    \frac{2^{n+1}+(-1)^n}{3\cdot 2^n}+\left(\frac{-1}{2}\right)^{n+1}=<br />
\frac{2^{n+1}+(-1)^n}{{\color{blue}3\cdot 2^n}}+\frac{(-1)^{n+1}}{{\color{red}2^{n+1}}}

    As 3\cdot 2^{n+1}=2\cdot {\color{blue}3\cdot2^n}=3\cdot{\color{red}2^{n+1}}, we choose 3\cdot 2^{n+1} as the common denominator. To make appear 3\cdot 2^{n+1} at the denominator of the first fraction, we multiply it by \frac{2}{2}. For a similar reason, we multiply the second fraction by \frac{3}{3}.


    Otherwise, one can do this as follows :

    \begin{aligned}<br />
\frac{2^{n+1}+(-1)^n}{3\cdot 2^n}+\left(\frac{-1}{2}\right)^{n+1}&=<br />
\frac{2^{n+1}+(-1)^n}{{3\cdot 2^n}}+\frac{(-1)^{n+1}}{{2^{n+1}}}\\<br />
&=\frac{1}{2^n}\left(\frac{2^{n+1}+(-1)^n}{3}+\frac{(-1)^{n+1}}{2}\right)\\<br />
&=  \frac{1}{2^n}\cdot\frac{2\left(2^{n+1}+(-1)^n\right)+3(-1)^{n+1}}{3\cdot 2}\\<br />
&=\ldots<br />
\end{aligned}<br />
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