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Thread: Another sums question =)

  1. #1
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    Another sums question =)

    Some sums questions again:
    Have some thoughts about if these solutions are correct..all comments highly appreciated
    First one out is:
    Write without Sigma notation
    $\displaystyle
    \sum_{k=1}^{n}\frac{k}{n+k}$

    well here is my shot should be:
    $\displaystyle \frac{1}{n+1}+\frac{2}{n+2} +\frac{3}{n+3}$




    Then we have a "calc the sum of stuff":
    $\displaystyle 3+7+11+15+...+43$
    well I figured since there is a 4 between each term why dont we: $\displaystyle (43+3)\times \frac{43-3}{4}$$\displaystyle = 460$






    We have another question: calculate the sum of:$\displaystyle 1+3+5+...+(2n-1)$
    well if I write it so the first number overlapswith the last
    1+ .............3+ ...........5...........+(2n-1)
    (2n-1)+1 +(2(n-1))+3 + (2(n-2))+5 + (2n-1)+1
    then $\displaystyle \frac{2n\times n}{2} = \frac{2n^2}{2} = n^2$





    Then a more difficult one calculate the sum of:
    $\displaystyle a-ak+ak^2-ak^3+...+a(-k)^n$



    $\displaystyle a(1-k(1-k)-k^3(1-k)-k^{n-1}(1-k))$ hey lets divide by a
    $\displaystyle a(1-k)(1-k-k^3-k^5-k^{n-1})$ oohh look divide by (1-k)
    $\displaystyle (a-ak)(1-k-k^3-k^5-k^{n-1}-k^n)$ hej lets add
    $\displaystyle (a-ak)(1-(n^2-k)) = a- an^2-2ak+ak^2+akn^2$ turned out like this.
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  2. #2
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    Quote Originally Posted by Henryt999 View Post
    Some sums questions again:
    Have some thoughts about if these solutions are correct..all comments highly appreciated
    First one out is:
    Write without Sigma notation
    $\displaystyle
    \sum_{k=1}^{n}\frac{k}{n+k}$

    well here is my shot should be:
    $\displaystyle \frac{1}{n+1}+\frac{2}{n+2} +\frac{3}{n+3}$


    Don't understand this one: you wrote the first three summands of the above series...so?


    Then we have a "calc the sum of stuff":
    $\displaystyle 3+7+11+15+...+43$
    well I figured since there is a 4 between each term why dont we: $\displaystyle (43+3)\times \frac{43-3}{4}$$\displaystyle = 460$


    This is the sum of an arithmetic series with first element 3 and common difference 4. If $\displaystyle a_1, a_2,\ldots,a_n$ is an arithmetic series, with common difference $\displaystyle d=a_n-a_{n-1}\,\,\forall n$ , then the sum of its first consecutive n terms is $\displaystyle S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$ .



    We have another question: calculate the sum of:$\displaystyle 1+3+5+...+(2n-1)$
    well if I write it so the first number overlapswith the last
    1+ .............3+ ...........5...........+(2n-1)
    (2n-1)+1 +(2(n-1))+3 + (2(n-2))+5 + (2n-1)+1
    then $\displaystyle \frac{2n\times n}{2} = \frac{2n^2}{2} = n^2$



    Correct. Again, this is the sum of an arithmetic series with difference 2 and first element 1


    Then a more difficult one calculate the sum of:
    $\displaystyle a-ak+ak^2-ak^3+...+a(-k)^n$



    $\displaystyle a(1-k(1-k)-k^3(1-k)-k^{n-1}(1-k))$ hey lets divide by a
    $\displaystyle a(1-k)(1-k-k^3-k^5-k^{n-1})$ oohh look divide by (1-k)
    $\displaystyle (a-ak)(1-k-k^3-k^5-k^{n-1}-k^n)$ hej lets add
    $\displaystyle (a-ak)(1-(n^2-k)) = a- an^2-2ak+ak^2+akn^2$ turned out like this.

    The sume of the first n consecutive elements of a geometric series $\displaystyle a_1,a_2,\ldots,a_n$ with common quotient $\displaystyle q=\frac{a_n}{a_{n-1}}\,\,\forall n$ is given by $\displaystyle S_n=a_1\frac{q^n-1}{q-1}$

    Here you have $\displaystyle a_1=1\,,\,\,q=-k$ so now just substitute.

    Tonio
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  3. #3
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    Dear Tonio

    Thank you for your help!
    The first exercise was menth to "show that I understand" what a Sigma notation menth.
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