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Math Help - Calculate the summations

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    Calculate the summations

    Calculate the summations
    Attached Thumbnails Attached Thumbnails Calculate the summations-codecogseqn-15-.gif   Calculate the summations-codecogseqn-14-.gif   Calculate the summations-codecogseqn-13-.gif   Calculate the summations-codecogseqn-12-.gif  
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    Re: Calculate the summations

    How far have you been able to work these? And are these 4 questions?

    -Dan
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    Re: Calculate the summations

    the firs problem
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    Re: Calculate the summations

    the second problem
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    Re: Calculate the summations

    the third problem
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    Re: Calculate the summations

    the last problem
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    Re: Calculate the summations

    till the farther far
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    Re: Calculate the summations

    i would also intressted on how to solve this
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    Re: Calculate the summations

    Quote Originally Posted by WORLD View Post
    the firs problem
    \Sigma _{k = 0} ^{100} \frac{1}{k + 1} \left ( _k ^{100} \right )

    What does \left ( _k ^{100} \right ) mean?

    -Dan
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    Re: Calculate the summations

    Itís mean
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    Re: Calculate the summations

    could any one help me
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    Re: Calculate the summations

    Quote Originally Posted by WORLD View Post
    the firs problem
    Find \sum_{k=0}^{100}\frac{1}{k+1} \binom{100}{k}^2
    Hello, WORLD!

    This is a tough problem for the algebra forum. I'm going to use a little calculus. I hope you will still find my remarks useful.

    Start with the Binomial Theorem:
    (1+x)^n = \sum_{i=0}^n \binom{n}{i} x^i

    Integrate both sides from 0 to x (this is the calculus part). The result is
    \frac{1}{n+1}[(1+x)^{n+1}-1] = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i} x^{i+1}

    Now multiply these two equations together. I'm going to replace i with j in the first equation, so it will be easier to keep everything straight.
    (*).... (1+x)^n \cdot \frac{1}{n+1}[(1+x)^{n+1}-1] = \sum_{j=0}^n \binom{n}{j} x^j \cdot \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i} x^{i+1}

    Now consider the coefficient of x^{n+1} in (*). The left-hand side of (*) is equal to
    \frac{1}{n+1}[(1+x)^{2n+1} - (1+x)^{n}] = \frac{1}{n+1} \left[ \sum_{i=0}^{2n+1} \binom{2n+1}{i} x^i - \sum_{i=0}^n \binom{n}{i}x^i \right]
    so the coefficient of x^{n+1} is
    \frac{1}{n+1} \binom{2n+1}{n+1}

    On the right-hand side of (*), the coefficient of x^{n+1} occurs when j = n-i, so the coefficient is
    \sum_{i=0}^n \binom{n}{n-i} \cdot \frac{1}{i+1} \binom{n}{i} = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i}^2
    since \binom{n}{i} = \binom{n}{n-i}.

    The coefficient of x^{n+1} must be the same on both sides of the equation, so we have
    \frac{1}{n+1} \binom{2n+1}{n+1} = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i}^2

    Now set n = 100, and you're done.
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