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Rules of Summation Help Me Please

  1. Nov 30, 2004 #1
    Rules of Summation....Help Me Please

    I have a vauge idea of what these rules mean:
    1. [tex]\sum^n_{i=1}c=cn[/tex]

    2. [tex]\sum^n_{i=1}i=\frac{n(n+1)}_{2}[/tex]

    3. [tex]\sum^n_{i=1}i^2=\frac{n(n+1)(2n+1)}_{6}[/tex]

    4. [tex]\sum^n_{i=1}i^3=\frac{n^2(n+1)^2}_4[/tex]

    are these rules saying that if i have an expression, (i) raised to a power from 1 to 3 then i use the above rules? How would you solve the following problem?

    [tex]\sum^n_{i=1}\frac{(i + 1)}_{n^2}[/tex] for n = 10,100,1000,10000

    i know that the answer reduces to [tex]\frac{(n + 3)}_{2n}[/tex]

    and you make some substitutions, but how is this done? any help would be appreciated. It's three am and i'm still trying to figure this out!
  2. jcsd
  3. Nov 30, 2004 #2
    [tex]\sum^n_{i=1}\frac{(i + 1)}{n^2} = \frac{1}{n^2}\left( \sum^n_{i=1}i + \sum^n_{i=1} 1 \right) = \frac{1}{n^2}\left( \frac{n(n+1)}{2} + n \right) = \frac{1}{n^2} \left( \frac{n^2 + 3n}{2} \right) = \frac{n + 3}{2n} [/tex]

    Once you have that, just plug the value for n.

  4. Nov 30, 2004 #3


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    Rewrite your sum as

    [tex]\sum_{i=1}^n \frac {i+1}{n^2} = \frac {1}{n^2} \sum_{i=2}^{n+1} i[/tex]

    and deduce the value of the summation from your Eq. 2.
  5. Nov 30, 2004 #4
    The rewriting part is the part that i don't undersatand. Are we solving these for n, or i?
  6. Nov 30, 2004 #5

    James R

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    i is the summation variable. For example, consider your equation:


    Suppose we choose n=3. Then, the sum can be written out term-by-term. The left hand side says:

    [tex]\sum^3_{i=1}i^2= 1^2 + 2^2 + 3^2 = 14[/tex]

    The right hand side says:

    [tex]\frac{n(n+1)(2n+1)}{6} = \frac{3(3+1)(2.3 + 1)}{6} = 14[/tex]
  7. Nov 30, 2004 #6
    i'm still not sure about these. what about this one:

    [tex]\sum^4_{k=0}\frac{1}_{k^2 + 1}[/tex]

    i know the answer is 158/85. i also know that you could solve this one by keeping the summation in the same form and letting k = 0,1,2,3,4 succesively, but how would you use the rules of summaton to compute the result?
    i thought about doing this:

    [tex]\sum^4_{k=0}(k^2 + 1)^-1[/tex]

    but got stuck and gave up on the idea
  8. Nov 30, 2004 #7


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    For that example, none of the "rules" you gave apply. The best thing to do, especially since the sum only involves 5 terms, is to add them up!

    [tex]\sum^4_{k=0}\frac{1}_{k^2 + 1}}= 1+ \frac{1}{2}+\frac{1}{5}+\frac{1}{10}+\frac{1}{17}[/tex]
  9. Nov 30, 2004 #8
    I've been trying to find a solution for that example without having to add the terms up manually. Is it even possible?

    I tried to find the sum by the method of differences. Here's what I figured out:

    [tex]\sum_{k=0}^{4} (k^2 + 1)^{-1} = \sum_{k=0}^{4} \;\: \left[\sum_{n=0}^{\infty} (k^4)^{n} - \sum_{n=0}^{\infty} (k^2)^{2n+1}\right] = \sum_{k=0}^{4} \left(\frac{1}{1-k^4} - \frac{k^2}{1-k^4}\right)[/tex]

    But that didn't really help.
  10. Nov 30, 2004 #9
    The summation symbol is the most powerful mathematical symbol there is, so you need to be certain that you do understand its usage. Its usage must be consistent with the axioms of algebra. With that in mind, consider the first equation above:


    The left hand side of this equation is a "shorthand" for the following longer expression:

    C1+C2+C3+C4+C5+.... +Cn where C1=C2=C3=...=Cn=C

    Now, use the distributive law of algebra to factor out the C to get this:


    There are n number ones inside the parenthesis, so that you get this after factoring out the C:

    C times n = nc

    So that the LHS is equivalent to the RHS, by an application of the distributive law of algebra.

    Your number two is far more interesting.


    The LHS uses the summation symbol as a shorthand for the following longer expression:

    1+2+3+4+5+... +n

    Which is nothing more than the sum of the first n natural numbers. In order to understand the formula on the RHS, consider this:


    = 5*6

    So that 1+2+3+4+5 = 5*6/2

    So that in general:

    1+2+3+...+n = n*(n+1)/2

    Which is the RHS by ad hoc reasoning.

    Now, without deriving number three above, i challenge you to prove that it is true using mathematical induction. You will find that it is true, indeed the LHS is numerically equivalent to the RHS.

    All this having been said, consider evaluating the following expression:

    [tex]\sum^n_{i=1}(6i^2 -4i + 7)[/tex]

    Using the axioms of algebra, the previous expression is numerically equivalent to the following expression:

    [tex]\sum^n_{i=1}6i^2 - \sum^n_{i=1}4i + \sum^n_{i=1}7[/tex]

    To find the final expression, you will need to know the following, preferably a priori:

    1. [tex]\sum^n_{i=1}6i^2 = 6\sum^n_{i=1}i^2= n(n+1)(2n+1)[/tex]
    2. [tex]\sum^n_{i=1}4i = 4\sum^n_{i=1}i= 2(n)(n+1) [/tex]
    3. [tex] \sum^n_{i=1}7=7n[/tex]

    Your final formula will not have the dummy variable i in it, rather it will be a function of n, where n is an element of the natural numbers.

    Then if you want to find out what the sum is for n = 10,100,1000,10000, i suggest you buy a pocket calculator for $20, and do the final multiplication. To check your answer, write a computer program with the summation formula in it, so that the computer actually adds, and then wait to see if the answer after long addition, is equivalent to the answer you got on your calculator.

    Just so you are fully aware, the axioms of algebra are being used to write these equations. In other words, consider something like this:

    [tex]\sum^n_{i=1}6i^2 [/tex]

    The previous expression is a shorthand for the following expression:

    6(1^2)+6(2^2)+6(3^2)+... + 6(n^2)

    It is only after an application of the distributive law of algebra, that we can say the previous expression is numerically equivalent to the following 'different looking' expression:


    And then we can use the summation shorthand to justify saying that the previous expression is numerically equivalent to the following expression:


    So now let us consider your question, you want to tackle the following SUM:

    [tex]\sum^n_{i=1}\frac{(i + 1)}_{n^2}[/tex]

    The first thing to realize, is that as the indicial variable i is varying, the number n is remaining constant. Since it is a constant, you can factor it out of the summation symbol (this is justified by the distributive axiom of algebra), so that you can write the following equation:

    [tex]\sum^n_{i=1}\frac{(i + 1)}_{n^2} [/tex]= [tex] 1/n^2 \sum^n_{i=1}(i + 1)[/tex]

    And now you need to know this:

    [tex] \sum^n_{i=1}(i + 1) =\sum^n_{i=1}i + \sum^n_{i=1}1 = n(n+1)/2 + n[/tex]

    Now, get a common denominator of 2.

    In other words:

    n(n+1)/2 + n = n(n+1)/2+2n/2= [ n(n+1)+ 2n ]/2 = n[(n+1) +2]/2 = n(n+3)/2

    Now, you can pull everything together to write this:

    [tex]\sum^n_{i=1}\frac{(i + 1)}_{n^2} [/tex] = 1/n^2 * n(n+3)/2 = (n+3)/2n

    as desired.
    Last edited: Nov 30, 2004
  11. Nov 30, 2004 #10
    thanks for all the help, it's starting to sink in now.
  12. Dec 6, 2004 #11
    Hey um, Tempus Rex--what did you mean by the summation symbol being the most powerful? What do you mean--most important?---(i dunno...just curious)

    (btw, it is my favorite symbol, and I do use it in almost every problem i get or make up)

    But how is it "most powerful"?--is there some sort of rank, or what makes the sum so important--well, compared to other symbols? It's useful---but powerful?
    Last edited: Dec 6, 2004
  13. Dec 6, 2004 #12
    Symbols are just that. Whats important is the concept behind them.
    I dont think you can talk that way about symbols, they are just tools to solve problems, not methods or theory's.

    Is the sumation concept more important or more powerfull than the concept of lenght, set, order, ...????

    Sumation might be the oldest concept in math, and maybe thats what Tempus mean, but all concepts and their symbols have a place to be in mathematicall theory.

    For example, Riemann Sums are a fundamental concept in the development of integration theory, but they are just a pain to deal with if you are actually trying to integrate something, so what can you judge of that?

    As a concept, a Riemann Sum is fully needed to understand and develope some concepts like integrability, etc, but as a notation, as a simple *symbol* is practically useless. The [itex]\int[/itex] symbol mostly deals with this problem, but tells you nothing of the theory or the concept.

    I think sumation is one of the pilar concepts of math development, but at this point in time, it can also be seen as just a function. The level of abstraction that we have reached permits us to fully understand and generalize such concepts. This happens also with the concept of lenght, and a lot more mathematical ideas. Thats why there isnt a more important or most powerfull concept (or symbol) in my opinion, and you shouldnt think in math that way.
  14. Dec 9, 2004 #13
    i can continue the power sums by focusing on numbers that have digits increasing from left to right.
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