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Evaluting Summation of Mutiplied Terms

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Evaluate the sums: [itex]\sum^{n}_{i=1}ia^{i}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I'm assuming that because there's no limit or anything, the professor wants an equation. I know [itex]\sum^{n}_{i=1}i[/itex] is [itex]\frac{1}{2}n(n+1)[/itex] and [itex]\sum^{n}_{i=1}a^{i}[/itex] is [itex]\frac{a^{n+1} - 1}{a-1}-1[/itex] but how do you deal with them when multiplied? I can't separate it out like I can in addition. Do I use integrals?
     
    Last edited: Sep 14, 2011
  2. jcsd
  3. Sep 14, 2011 #2
    There are different ways of doing this. There is a way to do it using calculus. Think about letting a be a variable, and integrating to get another expression that you know how to evaluate.
     
  4. Sep 14, 2011 #3
    I'm sorry, I'm really lost.

    There's a relation between summation and integration somewhere that I can use to my advantage is what I got from your reply. It's been a while since I've done any Calculus.

    Is there a resource available that I can study this more in depth? All I have is an Algorithms textbook and the Appendix is rather sparse.
     
  5. Sep 14, 2011 #4
    *EDIT*

    I'm sorry, I realize a different version of the problem is easier to do with calculus. That's the sum of (i+1)a^i.
     
  6. Sep 14, 2011 #5
    Try to evaluate the sum of (i+1)a^i from i=1 to n first. Integrating (i+1)a^i with respect to i, you get a^(i+1). Now, you have n terms in the sum. The integral of the sum is simply the sum of the integrals of the individual terms, i.e. the sum of the terms a^(i+1) from i=1 to n. This you already know a formula for. How would you proceed from here?
     
  7. Sep 14, 2011 #6
    Just to clarify, the method I'm proposing is NOT treating the sum as an integral. Rather, it's to integrate the expression to yield an easier expression which you can evaluate. Then you would need to take the derivative of that to get back the desired result.

    I'll show you a different method as well, which doesn't use Calculus.

    Let S represent the sum S=a+2a^2+...+na^n. Multiplying by a, we get aS=a^2+2a^3+...na^(n+1). Subtracting the latter equation from the former, we get (1-a)S=a+a^2+a^3+...+a^n-na^(n+1). The right side consists of a summation you already know how to evaluate (a+a^2+...+a^n) plus an extra term. Now it's just a matter of algebra to find S.
     
    Last edited: Sep 14, 2011
  8. Sep 14, 2011 #7
    Brilliant!

    Lesse,

    [itex]S = a + 2a^2 + 3a^3 + \ldots + na^n[/itex]
    [itex]aS = a^2 + 2a^3 = 3a^4 + \ldots + na^{n+1}[/itex]

    Subtracting, I get:

    [itex]S(1 - a) = a + a^2 + a^3 + a^4 + \ldots + a^n + na^{n + 1}[/itex]
    [itex]S(1 - a) = \frac{a^{n+1} - 1}{a - 1} - 1 + na^{n + 1}[/itex]
    [itex]S = \frac{a^{n+1} - 1}{(a - 1)(1 - a)} - \frac{1}{1 - a} + \frac{na^{n + 1}}{1 - a}[/itex]
    [itex]S = \frac{a^{n+1} - 1 - (a - 1) + na^{n+1}(a - 1)}{(a - 1)(1 - a)}[/itex]
    [itex]S = \frac{a^{n+1} - a + na^{n+1}(a - 1)}{(a - 1)(1 - a)}[/itex]

    I guess I can't get any neater than that.

    EDIT: Whoops. Did it incorrectly.

    [itex]S(1 - a) = \frac{a^{n+1} - 1}{a - 1} - 1 - na^{n + 1}[/itex]
    [itex]S = \frac{a^{n+1} - 1}{(a - 1)(1 - a)} - \frac{1}{1 - a} - \frac{na^{n + 1}}{1 - a}[/itex]
    [itex]S = \frac{a^{n+1} - 1 - (a - 1) - na^{n+1}(a - 1)}{(a - 1)(1 - a)}[/itex]
    [itex]S = \frac{a^{n+1} - a - na^{n+1}(a - 1)}{(a - 1)(1 - a)}[/itex]
    [itex]S = \frac{a^{n+1} - a - na^{n+1}(a - 1)}{(a - 1)(1 - a)}[/itex]

    The other way you mentioned:

    Erm...

    [itex]\sum^n_{i=1}\int(i + 1)a^i di = \sum^n_{i=1}a^{i+1} = \frac{a^{n+1} - 1}{a - 1} - 1 - a + a^{n+1}[/itex]

    Then I'm suppose to take the derivative of that right?

    But how is [itex](i + 1)a^i[/itex] related to [itex]ia^i[/itex]?

    ... every term gets multiplied by +, how is that difference accounted for?

    Thank you for your patience.
     
    Last edited: Sep 14, 2011
  9. Sep 14, 2011 #8

    Char. Limit

    User Avatar
    Gold Member

    (i+1)a^i = i a^i + a^i

    Taking sums, we get something like this:

    [tex]\sum_{i=1}^n (i+1) a^i = \sum_{i=1}^n i a^i + \sum_{i=1}^n a^i[/tex]

    The first term is one that you can figure out, as it's the derivative of a^(i+1), and the derivative of a sum is the sum of the derivatives, so you can represent it as:

    [tex]\sum_{i=1}^n (i+1) a^i = \sum_{i=1}^n \left(\frac{d}{da} a^{i+1}\right) = \frac{d}{da} \left(\sum_{i=1}^n a^{i+1}\right)[/tex]

    The third term is also simple. Using those two, you can find the value of the second term.
     
  10. Sep 14, 2011 #9

    Ray Vickson

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    Science Advisor
    Homework Helper

    That is a standard method, but in this case you need to *differentiate*, not integrate. Think of i*a^i as a*(d/da)(a^i).

    RGV
     
  11. Sep 14, 2011 #10
    Oh! I see. So that was what he meant.

    Thanks.
     
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