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Calculus for finite sum

  1. May 26, 2007 #1
    Is there some way that using calculus we can explicitely sum up :
    [tex]\sum_{i=0}^{n}(i+1)^2 \binom{n}{i}=[/tex]

    For example,applying the derivative to the both sides of expression
    we quickly get how much is [itex]1+2x+3x^2+...+nx^{n-1}=[/itex].
    But I don't know what to do with the first sum.Tryed binomial theorem and some identies but couldn't make use of them.
    Thanks for your advice in advance.
    Last edited: May 26, 2007
  2. jcsd
  3. May 28, 2007 #2
    You can use this ideas:


    [tex]i^2\times\frac{n!}{i!(n-i)!}=i\times\frac{i\times n!}{i!(n-i)!}= [/tex]

    [tex]i\times\frac{n\times (n-1)!}{(i-1)!(n-i)!}=
    i\times\frac{n\times (n-1)!}{(i-1)!(n-i)!}=[/tex]

    [tex]n\times\frac{(i-1+1)\times (n-1)!}{(i-1)!(n-i)!}=[/tex]
    [tex]n\times\frac{(n-1)!}{(i-1)!(n-i)!}+n\times\frac{(i-1)\times (n-1)!}{(i-1)!(n-i)!}=[/tex]

    [tex]n\times\frac{(n-1)!}{(i-1)!(n-i)!}+n\times\frac{(n-1)\times (n-2)!}{(i-2)!(n-i)!}[/tex]
    Last edited: May 28, 2007
  4. May 30, 2007 #3
    Well I tryed and I couldn't make much of progress with this.
    This problem I found in one of my calculus textbooks and I think we are supposed to use calculus methods somehow...
  5. May 31, 2007 #4


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    I think you can do the sum like this

    Multiply both sides by x
    Differentiate both sides wrtx
    Multiply both sides by x
    Differentiate both sides wrtx
    Let x=1, after some simplification,
    Last edited: May 31, 2007
  6. May 31, 2007 #5
    Thanks!I knew of the result but didn't know how to derive it.
  7. Jun 1, 2007 #6
    Your solution gave me idea to find another solution with calculus method.
    Basically the two are equivavalent but I played with antiderivatives.

    If [tex]f(x)=1^2 + 2^2\binom{n}{1}x + 3^2\binom{n}{2}x^2 + ...+(n+1)^2\binom{n}{n}x^n[/tex]

    Then [tex]F(x)=\int_{0}^{x}f(t)dt=x+2\binom{n}{1}x +3\binom{n}{2}x^2+...+(n+1)\binom{n}{n}x^n[/tex]

    [tex]g(x)=\frac{F(x)}{x}=1+2\binom{n}{1}x + 3\binom{n}{2}x^2+...+(n+1)\binom{n}{n}x^n[/tex]

    Integrating one more time:

    Now we conclude:
    [tex]F(x)=x(1+x)^n +nx^2(1+x)^{n-1}[/tex]
    [tex]f(x)=(1+x)^n +3nx(1+x)^{n-1}+n(n-1)x^2(1+x)^{n-2}[/tex]

    Substitution x=1 in the last one gives the result.
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