Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook