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Math Help - Binomial expansion

  1. #1
    Member srirahulan's Avatar
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    Lightbulb Binomial expansion

    If S_{n}=\frac{^{n}c_{1}}{1}-\frac{^{n}c_{2}}{2}+\frac{^{n}c_{3}}{3}-\frac{^{n}c_{4}}{4}+..........+\frac{(-1)^{n-1} \ ^{n}c_{n}}{n}

    Show that
    S_{n+1}-S_{n}=\frac{1}{n+1}

    I go by this way>>>>

    n----> n+1

    S_{n+1}=\frac{^{n+1}c_{1}}{1}-\frac{^{n+1}c_{2}}{2}+\frac{^{n+1}c_{3}}{3}-\frac{^{n+1}c_{4}}{4}+..........+(-1)^{n-1}\frac{^{n-1}c_{n}}{n}+(-1)^{n}\frac{^{n+1}c_{n+1}}{n+1}
    After that
     S_{n+1}=\frac{^{n}c_{1}+^{n}c_{0}}{1}-\frac{^{n}c_{2}+^{n}c_{1}}{2}+\frac{^{n}c_{3}+^{n}  c_{2}}{3}-\frac{^{n}c_{4}+^{n}c_{3}}{4}+...................+  (-1)^{n-1}\frac{^{n-1}c_{n}+^{n-1}c_{n-1}}{1}+(-1)^{n}\frac{^{n}c_{n+1}+^{n}c_{n}}{n+1}
    But when i subtract i could not get like that>>>>>
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  2. #2
    MHF Contributor
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    Re: Binomial expansion

    Your notation is slightly confusing. The title of your post "Binomial Expansion" could suggest two possible ways to interpret your notation. You might be using {^n}C_r = \dfrac{n!}{(n-r)!r!}, which is a binomial coefficient, or you may be asking that we help you expand the following using Newton's Generalized Binomial Formula:

    {^nc} = \underbrace{c^{c^{.^{.^{.^c}}}}}_{n\text{ times}}

    I believe you mean the former, so I will change the notation slightly:

    S_n = \frac{{_n}C_1}{1} - \frac{{_n}C_2}{2} + \frac{{_n}C_3}{3} - \frac{{_n}C_4}{4} + \cdots + \frac{(-1)^{n-1}{_n}C_n}{n}

    where currently, we assume {_n}C_r is only defined when n,r are nonnegative integers. Then {_n}C_r = \begin{cases}\dfrac{n!}{r!(n-r)!} & \text{if }n\ge r \\ 0 & \text{otherwise}\end{cases}

    So, S_{n+1} = \sum_{r=1}^{n+1} (-1)^{r-1}\dfrac{{_{n+1}}C_r}{r} (as you wrote in your post).

    However, let's manipulate this a little before we subtract S_n. Using Pascal's Rule, which is {_{n+1}}C_r = {_n}C_{r-1}+{_n}C_r, we get:

    \begin{align*}S_{n+1} & = \sum_{r=1}^{n+1} (-1)^{r-1}\dfrac{{_{n+1}}C_r}{r} \\ & = \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1} + {_n}C_r}{r} \\ & = \left( \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1}}{r} \right) + \left( \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_r}{r} \right) \\ & = \left( \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1}}{r} \right) + \left( \sum_{r=1}^n (-1)^{r-1} \dfrac{{_n}C_r}{r} \right) + (-1)^{n+1-1} \dfrac{{_n}C_{n+1}}{n+1} \\ & = \left( \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1}}{r} \right) + \left( \sum_{r=1}^n (-1)^{r-1} \dfrac{{_n}C_r}{r} \right) + (-1)^n \dfrac{0}{n+1} \\ & = \left( \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1}}{r} \right) + S_n\end{align*}

    Subtracting S_n from both sides gives:

    \begin{align*}S_{n+1} - S_n & = \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1}}{r} \\ & = \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{n!}{(n+1-r)!r!} \\ & = n!\sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{1}{(n+1-r)!r!} \\ & = n!\left(\dfrac{1}{n!1!} - \dfrac{1}{(n-1)!2!} + \cdots + (-1)^{n-2}\dfrac{1}{2!(n-1)!} + (-1)^{n-1}\dfrac{1}{1!n!} + (-1)^n\dfrac{1}{0!(n+1)!} \right) \end{align*}

    Let's look at this for small values of n.

    (n=1):
    S_2-S_1 = 1!\left(\dfrac{1}{1!1!} - \dfrac{1}{0!2!} \right) = \dfrac{1}{2} = \dfrac{1}{1+1}

    (n=2):
    S_3 - S_2 = 2!\left(\dfrac{1}{2!1!} - \dfrac{1}{1!2!} + \dfrac{1}{0!3!} \right) = 2!\left(0 + \dfrac{1}{3!}\right) = \dfrac{1}{3} = \dfrac{1}{2+1}

    (n=3):
    \begin{align*}S_4 - S_3 & = 3!\left(\dfrac{1}{3!1!} - \dfrac{1}{2!2!} + \dfrac{1}{1!3!} - \dfrac{1}{0!4!} \right) \\ & = 3!\left( 2\dfrac{1}{3!}-\dfrac{1}{2!2!}-\dfrac{1}{4!} \right) \\ & = 2-\dfrac{3}{2}-\dfrac{1}{4} = \dfrac{1}{4} = \dfrac{1}{3+1}\end{align*}

    (n=4):
    \begin{align*}S_5 - S_4 & = 4!\left( \dfrac{1}{4!1!}-\dfrac{1}{3!2!}+\dfrac{1}{2!3!} - \dfrac{1}{1!4!} + \dfrac{1}{0!5!} \right) \\ & = 4!\left(0+\dfrac{1}{5!}\right) = \dfrac{1}{5} = \dfrac{1}{4+1}\end{align*}

    The pattern is that when n is even, the first n terms cancel each other out. When n is odd, the first n terms should add to \dfrac{2}{n+1}.

    To see that this is true, suppose n is even. Then we have:

    \begin{align*}\sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{n!}{(n+1-r)!r!} & = \sum_{r=1}^{n/2}(-1)^{r-1}\dfrac{n!}{(n+1-r)!r!} + \sum_{r=n/2+1}^n (-1)^{r-1}\dfrac{n!}{(n+1-r)!r!} + (-1)^n\dfrac{n!}{0!(n+1)!}\\ & = \left(\sum_{r=1}^{n/2} (-1)^{r-1}\dfrac{n!}{(n+1-r)!r!} + \sum_{r=1}^{n/2}(-1)^{n-r}\dfrac{n!}{r!(n+1-r)!}\right) + \dfrac{1}{n+1}\end{align*}

    When r is odd, then r-1 is even and n-r is odd, so (-1)^{r-1}\dfrac{n!}{(n+1-r)!r!} + (-1)^{n-r}\dfrac{n!}{r!(n+1-r)!} = \dfrac{n!}{(n+1-r)!r!} - \dfrac{n!}{r!(n+1-r)!} = 0. When r is even, then r-1 is odd and n-r is even, so (-1)^{r-1}\dfrac{n!}{(n+1-r)!r!} + (-1)^{n-r}\dfrac{n!}{r!(n+1-r)!} = -\dfrac{n!}{(n+1-r)!r!}+\dfrac{n!}{r!(n+1-r)!} = 0. Hence, the only term that remains is the \dfrac{1}{n+1}.

    See if you can figure out how to show that when n is odd, you get \dfrac{2}{n+1}.
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  3. #3
    Member srirahulan's Avatar
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    Re: Binomial expansion

    Ohh That's my fault>>>Thank for your answer>>
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